IV Approximation of Rational Functions 1. IV.C Bounding (Rational) Functions on Intervals... 4

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1 Contents IV Approxiation of Rational Functions 1 IV.A Constant Approxiation IV.B Linear Approxiation IV.C Bounding (Rational) Functions on Intervals IV.D The Extree Value Theore (EVT) for Rational Functions

2 IV Approxiation of Rational Functions IV.1 IV Approxiation of Rational Functions Suppose f(x) is a rational function. An approxiating function, A(x), is another function usually uch sipler than f whose values are approxiately the sae as the values of f. The difference f(x) A(x) = E(x) is the error, and easures how accurate or inaccurate our approxiation is. Thus f(x) = A(x) + E(x) (approxiation) (error) The two siplest functions are constant functions and linear functions and if A(x) is constant or linear we talk about a constant approxiation, or a linear approxiation for f. When we choose our approxiation we want the error to be sall. Now we know what it eans for a nuber to be sall, but our error, E(x), is a function and it is less clear what it eans for a function to be sall. In fact there are different notions of sallness for functions and depending on which one uses, one gets different kinds of approxiations. In this section we shall fix a doain (f) and we shall say our error is sall if E(x) is very sall for all x very close to a. Thus our approxiation will be good if A(x) is very close to f(x) for all x very close to a. Accordingly, we want to find an approxiation for a rational function f(x) on a sall interval [a h, a+h] do(f) by soe sipler function A(x). Then we want to deterine the accuracy of the approxiation by bounding the absolute value of the error E(x) = f(x) A(x). In order to use the approxiation we need to have an idea of how big the error is. (If the error is 5 ties the value of the function, the approxiation is clearly no good!) The MVT and EMVT allow us to get a bound on the error. IV.A Constant Approxiation The best constant approxiation of f is A 0 (x) = f(a). graph of f(x) E0( x) f(a) (a, f( a)) f(x ) graph of A 0 (x) A 0 (x ) a x We write the error ter as E 0 (x) = f(x) A 0 (x) = f(x) f(a). The MVT tells us that f (z) = f(x) f(a) x a for soe z between a and x. Therefore f (z)(x a) = f(x) f(a) = E 0 (x).

3 IV.A Constant Approxiation IV. So to bound the error on soe interval [a h, a + h] around a, we need to bound E 0 (x) = f (z) x a on [a h, a + h]. Observe that x a h and so E 0 (x) ax f (z) h. We can thus bound the error z E 0 (x) as soon as we can bound f (z) on [a h, a + h]. Exaple. Let f(x) = x 3 4x + 5 and a =. Find A 0 (x) and a nuber h so that E 0 (x) 1 0 = δ when h x + h. Solution: A 0 (x) = f() = 5. How to find h when given a and δ? 1. Choose h 0 so that [a h 0, a + h 0 ] do(f). (h 0 is an initial guess for h). Find M so that f (z) M on [a h 0, a + h 0 ]. 3. Choose h in (0, h 0 ] so that M h δ. (Reeber we want h > 0, thus the ( in (0, h 0 ].) If we succeed with 1. 3., we get: If x [a h, a + h], then E 0 (x) = f (z) x a Mh δ. iplications a h x a + h 0<h h 0 Suarized as a diagra we have the z between x and a a h 0 x a + h 0 a h 0 z a + h 0 h x a h as M is upper bound for f (z) on [a h 0, a + h 0 ] f (z) M as 0 x a h and Mh δ x a h E 0 (x) = f (z) x a Mh δ and it is the botto iplication we want to achieve. Coing back to the actual exaple: 1. Choose h 0 = 1, then [a h 0, a + h 0 ] = [1.9,.1].. We want to bound the absolute value of f (z) = 3z 4: For z [1.9,.1], we have 1.9 z.1 = 3.61 z 4.41 =.83 3z 13.3 = z Thus f (z) = 3z = M on [1.9,.1]. 3. Find h in (0, 1 1 ] so that 9.3 h = Mh δ = We want 0 < h 1 Let h =.005. Fro 1.,., 3. we get finally To spell it out: 0. = 0.1 and h = h = x.005 = + h = E 0 (x) M h = = 1 0.

4 IV.B Linear Approxiation IV.3 The constant approxiation A 0 (x) = 5 to f(x) = x 3 4x + 5 is accurate to within an error E 0 (x) of at ost 1 0 as long as x [ 0.05, ]. IV.B Linear Approxiation The best linear approxiation for f near a is A 1 (x) = f(a) + f (a)(x a) the function whose graph is the tangent line to the graph of f through (a, f(a)). It is called the linear approxiation to f at a and its error is E 1 (x) = f(x) f(a) f (a)(x a) graph of A 1 (x) E 1 ( x) graph of f(x) (a, f(a)) A 1 ( x) f(x ) a x The EMVT tells us that f (z)(x a) = f(x) f(a) f (a)(x a) = f(x) A 1 (x) = E 1 (x), for soe z between a and x. Therefore we can control the error E 1 (x) as soon as we can find an upper bound, f (z) M, for the second derivative. Exaple. Let f(x) = x 3 4x + 5 and a =. Find A 1 (x) and a nuber h so that E 1 (x) 1 0 = δ when h x + h. Solution. A 1 (x) = f() + f ()(x ) = 5 + 8(x ) = 8x 11. How to find h? 1. Choose h 0 so that [a h 0, a + h 0 ] do(f). (h 0 is again an initial guess for h.). Find M so that f (z) M on [a h 0, a + h 0 ]. 3. Choose h in (0, h 0 ] so that Mh δ. If we succeed with 1. 3., we obtain: If x [a h, a + h], then E 1 (x) = f (z) (x a) Mh δ.

5 IV.C Bounding (Rational) Functions on Intervals IV.4 In our exaple: 1. Choose h = 1 ; [a h, a + h] = [1.9,.1] (the sae guess as above why not?).. We want to bound the absolute value of f (z) = 6z: For z [1.9,.1], we have 1.9 z.1 = f (z) 6(.1) = 1.6 = M Thus f (z) 1.6 = M on [1.9,.1]. 3. Find h in (0, 0.1] so that 1.6 h = Mh δ = 1 0. Therefore we want 0 < h 0.1 and h δ M = = or h Let h = We get finally that for h = 1.9 x.08 = + h = E 1 (x) 1 0. If you copare this linear approxiation with the constant one, to achieve the sae accuracy the interval has widened fro [1.995,.005] for the constant approxiation to [1.9,.08] for the linear approxiation. In other words, the linear approxiation is better than the constant approxiation as should be expected. Indeed, if f (z) M 1 and f (z) M on soe interval [a h, a + h], then on that interval E 0 (x) M 1 x a and E 1 (x) M x a. Both E 0 (x), E 1 (x) go to 0 as x a, but E 1 (x) goes uch faster than E 0 (x): x a 1/ 1/0 1/00 1/000 Bound for E 0 M 1 / M 1 /0 M 1 /00 M 1 /000 Bound for E 1 M /00 M /0000 M / M / IV.C Bounding (Rational) Functions on Intervals In order to ake the preceding ethods work, we need soe way of getting the constants M 1 and M. These are to be chosen so that f (z) M 1, z [a h, a + h] and f (z) M, z [a h, a + h]. As you will see below, we also want soeties to find a constant such that 0 < f (z), z [a h, a+h]. Since f and f are rational functions, we need soe way of solving the following. General Proble. M such that Given an interval [a, b] in the doain of a rational function g, find constants and g(x) M, for all x in [a, b]. Soe Techniques: 1. (Rules for absolute values; see also... ) If a, b R, then i) a b = a b

6 IV.C Bounding (Rational) Functions on Intervals IV.5 ii) a b = a b iii) a b = a n b n for every n 1. iv) a + b a + b (triangle inequality) v) a b a b (second version of the triangle inequality). Exaple. Find an upper bound for x 7 6x 4 + x on [, 1]. Solution: x [, 1] x 1 = x x, and for x one has x 7 6x 4 + x (iv) x 7 6x 4 + x (iv) x 7 + 6x 4 + x (i) = = x x 4 + x = x x 4 + x (iii) = 6.. (Bounds for reciprocals) 0 < n Q(x) N = 0 < 1 N 1 Q(x) 1 n. Exaple. Find an upper bound for 1 x 4 +x +6 on R. Solution: x 4 + x + 6 = x 4 + x = x 4 +x , for every x R. 3. (Bounding fractions) If P (x) M and 0 < n Q(x) N on [a, b], then N P (x) Q(x) M n every x [a, b]. Exaple. Find an upper bound for P (x) Q(x) = x7 6x 4 +x x 4 +x +6 on [, 1]. for Solution: P (x) Q(x) M n = 6 6 = (Critical points) If f(x) is a rational function defined on [a, b], then f (x) is defined on [a, b]. Furtherore, the extreal values that is, inia or axia of f(x) on [a, b] can occur only at those points where x = a, x = b, or f (x) = 0. Exaple. Bound f(x) = x 4 + x 3 x on [.5, 0]. f (x) = 4x 3 + 6x 4x = (x + )x(x 1). Therefore the roots of f (x) are, 0, 1. We don t care about x = 1 as it is not in the interval we are interested in. At the reaining points x =.5, x = 0 (the endpoints) and x = (the only root of f in (.5, 0)), we need to evaluate f. We find f(.5) = f(0) = 0 f( ) = 8 Therefore 8 f(x) 0 on [.5, 0], and so 0 f(x) 8 on [.5, 0]. 5. (Cobined ethods) To bound a rational function f(x) = P (x) Q(x), one can cobine the above ethods. For exaple, we can find a lower bound for f(x) by finding a lower bound for P (x) and an upper bound N for Q(x) using the ethod of critical points (3.) for each of P and Q and then using (4.) to bound the fraction f(x) fro below by N. Alternatively, one ight use (1.) for P and (3.) for Q or whatever cobination is suitable. Exaple. It would be foolish to try to bound f(x) = x 7 6x 4 +x x 4 +x +6 fro above by exaining the roots of f (x) if you don t believe it, calculate f (x)! Indeed, the solution given in (3.) above is optial here. Iportant Notes:

7 IV.C Bounding (Rational) Functions on Intervals IV.6 If one wants to bound f(x) on an interval that is not closed, say on an open interval (a, b), it is of course sufficient to bound the function on [a, b] as f(x) M for x [a, b] = f(x) M for x (a, b). You just have to ake sure that f is not only defined on (a, b) but on [a, b] as well. Make sure the function is defined where you want to bound it! Otherwise you ight end up with an arguent like 1 x 1 for x [ 1, 1] as (?) 1 1 = 1 1 = 1 and ( ) 1 x = 1 x is never zero (??). Checking the values of f at the endpoints, x = a, x = b, and at the roots of f on (a, b), tells you what the iniu/axiu of f on [a, b] is but not (directly) what the iniu/axiu of f on [a, b] is! That has to be established separately. Exaple. Assue we know that f is a rational function defined on [a, b], satisfying 8 f(x) for x [a, b], and taking on both extreal values 8 and on [a, b]. Then we know by the IVT that f(x) has a zero in [a, b] and we get on [a, b]: Miniu Maxiu f(x) 8 f(x) 0 8 A istake that happens all too often is of the following sort: Yikes! 8 f(x) = 8 f(x). It is crucial to reeber on which interval you want to bound a function! Exaple. Find lower and upper bounds for f(x) = x 3 9x + 4 on (i) [ 3, 3], (ii) [ 1, 1], (iii) [1, 3], (vi) ( 5, 4), (v) [ 3, 1). What is an upper bound for f on these intervals? Solution: The derivative of f(x) is f (x) = 3x 9 = 3(x 3)(x + 3). Evaluating f at the various endpoints and at the roots ± 3 of f, we find the following table where we check ( ) those points relevant to the interval in question and finally read off a lower () and upper bound (M): x f(x) [ 3, 3] [ 1, 1] [1, 3] ( 5, 4) [ 3, 1) M M 1 1 M M 4 3 M lower bound for f(x) upper bound upper bound for f

8 IV.D The Extree Value Theore (EVT) for Rational Functions IV.7 As a picture tells ore than a thousand words: (a) f(x)=x^3-9x (c) f(x) on [-1,1] 15 M (e) f(x) on (-5,4) 40 M (b) f(x) on [-3,3] 15 M (d) f(x) on [1,3] 4 M (f) f(x) on [-3,-1) M IV.D The Extree Value Theore (EVT) for Rational Functions We have just discussed techniques for finding bounds. Are there always such bounds? The answer is no : On (0, 1), the rational function f(x) = x 1/ x(x 1) has neither a lower nor an upper bound. If x 0 fro inside the interval (0, 1), then f(x) +, whereas if x 1 fro inside the interval (0, 1), then f(x). The situation is fortunately different for closed intervals: The Extree Value Theore (EVT for rational functions): Let f be a rational function defined on [a, b]. Then f takes on an absolute iniu and an absolute axiu M. There are thus c, d [a, b] such that f(c) = and f(d) = M, and f(x) M for all x [a, b].

9 IV.D The Extree Value Theore (EVT) for Rational Functions IV.8 Note: There ay be several arguents z [a, b] such that f(z) = and there ay be as well several arguents u [a, b] such that f(u) = M. The absolute iniu or axiu ay be taken on at one of the endpoints. The EVT akes no prediction at which arguent in [a, b] the function will take on either iniu or axiu. Cobining the IVT and the EVT, one obtains the following stateent. Corollary: If f is defined on [a, b], the iage of this closed interval under f is again a closed interval, f ([a, b]) = [, M]. The nuber is the absolute iniu of f on [a, b], the nuber M is the absolute axiu of f on [a, b]. Proof: The EVT says that there are, M R such that f ([a, b]) [, M] and that and M are taken on as values. The IVT says that any value between and M is taken on as well. Thus there are no gaps and f ([a, b]) = [, M].

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