The expansion of a function into a Taylor (or Maclaurin) series is useful as an approximation device in the circumstance that ^ 0 as n oc, but our present concern is with its application in the development of a general test for a relative extremum.

Taylor Expansion and Relative Extremum

As a preparatory step for that task, let us redefine a relative extremum as follows:

A function fix) attains a relative maximum (minimum) value at v0 jf f(x) - /(.r<j) is negative (positive) for values ofx in the immediate neighborhood of both to its left and to its right.

This can be made clear by reference to Fig. 9,10, where X} is a value of x to the left of.ro, and a"2 is a value of x to the right of.ro- In Fig. 9AQa? /(jcy) is a relative maximum; thus f{X[)) exceeds both /{*i) and /(x2). In short, fix) - f(x$) is negative for any value of.t in the immediate neighborhood of xq. The opposite is true of Fig. 9,10k where f(x^) is a relative minimum, and thus f(x) - f(xo) > 0.

Assuming f(x) to have finite, continuous derivatives up to the desired order at the point x =jcc> the function f{x)—not necessarily polynomial—can be expanded around the point xa as a Taylor series. On the basis of (9.14) (after duly changing <p to/), and using the Lagrange form of the remainder, wc can write

/(*) - fix o) = fix ,){x - Xo) + - xo)2 + ■■■

FIGURE 9.10

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MX |
M)< | |||||

L |
L------ |
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If the sign of the expression f(x) - /(*o) can be determined for values ofx to the immediate left and right of xq, we can readily come to a conclusion as to whether /(*o) is an extremum, and if so, whether it is a maximum or a minimum. For this, it is necessary to examine the right-hand sum of (9.17). Altogether, there are (n + 1) terms in this sum—n terms from Pn, plus the remainder which is in the (n + l)st degree—and thus the actual number of terms is indefinite, being dependent upon the value of n we choose. However, by properly choosing n, we can always make sure that there will exist only a single term on the right. This will drastically simplify the task of evaluating the sign of f(x) - fix») and ascertaining whether f(xo) is an extremum, and if so, which kind. Some Specific Cases This can be made clearer through some specific illustrations. If the first derivative at is nonzero, let us choose n = 0, so that the remainder will be in the first degree. Then there will be only « + 1 = 1 term on the right side, implying that only the remainder will be there. That is, we have where p is some number between xq and a value ofjc in the immediate neighborhood ofxo. Note that p must accordingly be very, very close to , What is the sign of the expression on the right? Because of the continuity of the derivative, f\p) will have the same sign as f{x o) since, as mentioned before, p is veiy, very close to In the present case, fip) must be nonzero; in fact, it must be a specific positive or negative number. But what about the ix — xq) part? When we go from the left of xo to its right, x shifts from a magnitude X\ < xo to a magnitude x2 > xq (sec Fig. 9.10). Consequently, the expression (x - xq) must turn from negative to positive as we move, and /to - /to) - f(P)(x ~ *o) must also change sign from the left of to its right. However, this violates our new definition of a relative extreinum; accordingly, there cannot exist a relative extremum at f(x o) when fix o) ^ 0 -a fact that is already well known to us. In this ease, choose n = 1, so that the remainder will be in the second degree. Then initially there will be n -I-1 = 2 terms on the right. But one of these terms will vanish because /'(xo) = 0, and we shall again be left with only one term to evaluate: As before, fip) will have the same sign as fix 0), a sign that is specified and unvarying, whereas the (x — Xo)2 part, being a square, is invariably positive. Thus the expression fix) - fix o) must take the same sign as /"(xq) and, according to the earlier definition of relative extremum, will specify You wili recognize this as the second-derivative test introduced earlier. Here we are encountering a situation that the second-derivative test is incapable of handling, for /"(xo) is now 7ero- With the help of the Taylor series, however, a conclusive result can be established without difficulty. Let us choose n = 2; then three terms will initially appear on the right. But two of these will drop out because fix 0) = /"to) = 0, so that we again have only one term to evaluate: fix) - fix o) = /(*„)(* - xo) + ^f"(x,)(x - xo)2 + - xo)J = \r\p)(.x - y,Y [because fix,) = 0, f"(x0) = 0] 0 As previously the sign of ff(p) is identical with that of /"'to) because of the continuity of the derivative and because p is very close to x0. But the (x — xo)3 part has a varying sign. Specifically, since (x - xn) is negative to the left of xo, so also will be (x ~ x0)3; yet, to the right of xo, the (x ^xo)3 part W»H be positive. Thus there is a change in the sign of f(x) - fix o) as we pass through xo, which violates the definition of a relative extremum. However, we know that xo is a critical value [/'to) = 0], and thus it must give an inflection point, inasmuch as it docs not give a relative extremum. Case 4 fix,) = fix,) = - - - = f*~]\x 0) - 0, but/< v>(*o) + 0 This is a very general case, and we can therefore derive a general result from it. Note that here all the derivative values are zero until we arrive at the Mh one. Analogously to the preceding three cases, the Taylor series for Case 4 will reduce to f(x)-f(x0) = ^JiN\p){x-x0f Again, takes the same sign as /*-V*(jco)> which is unvarying. The sign of the (x - xo)** part, on the other hand, will vary if N is odd (cf. Cases 1 and 3) and will remain unchanged (positive) if N is even (cf. Case 2). When N is odd» accordingly, f(x)- /(x0) will change sign as we pass through the point x0> thereby violating the definition of a relative extremum (which means thatxo must give us an inflection point on the curve). But when Nis even, f(x) - /(x0) will not change sign from the left of x0 to its right, and this will establish the stationary value /(x0) as a relative maximum or minimum, depending on whether is negative or positive, Nth-Derivative Test At last, then, we may state the following general test, Mh-Derivative test for relative extremum of a function of one variable If the first derivative of a function /(x) atxrt is /'(x0) = 0 and if the first nomero derivative value at xo encountered in successive derivation is that of the Mh derivative, f^\xo) ^ 0, then the stationary value /(xo) will be a. A relative maximum if N is an even number and /{jV)(xo) < 0. bt A relative minimum if N is an even number but /(jV\xo) > 0. c. An inflection point if N is odd. It should be clear from the preceding statement that the Mh-derivative test can work if and only if the function f(x) is capable of yielding, sooner or later, a nonzero derivative value at the critical value xo. While there do exist exceptional functions that fail to satisfy this condition, most of the functions we are likely to encounter will indeed produce nonzero in successive differentiation} Thus the test should prove serviceable in most instances. f If f(x) is a constant function, tor instance, then obviously f'{x) = f"{x) = ... = 0, so that no nonzero derivative value can ever be found- This, however, is a trivial case, since a constant function requires no test for extremum anyway. As a nontrivial exampfe, consider the function where the function y = >*? is an exponential function, yet to be introduced (Chap. 10). By itself, y = is discontinuous at x - 0, because x = 0 is not in the domain (division by zero is undefined). However, since lim y = 0, we can, by appending the stipulation that y = 0 for x = 0, fill X—* o the gap in the domain and thereby obtain a continuous function. The graph of this function shows that it attains a minimum at x = 0. But it turns out that, at x = 0, all the derivatives (up to any order) have zero values. Thus we are unable to apply the Nth-derivative test to confirm the graphically ascertainable fact that the function has a minimum at x = 0. For further discussion of this exceptional case, see R. Courant, Differential and Integral Calculus (translated by E.). McShane), Interscience, New York, vol. I, 2d ed., 1937, pp. 196, 197, and 336. Example 1 Examine the function y = (7 - x)4 for its relative externum, Since f '{x) = -4(7 - x)3 is --- zero when x = 7, we take x = 7 as the critical value for testing, with y = 0 as the stationary value of the function. By successive derivation (continued until we encounter a nonzero derivative value at the point x = 7), wc get f"(x) = 12(7 - xf so that r[7) = 0 Since 4 is an even number and since fH)(7) is positive, we conclude that the point (7, 0) represents a relative minimum. As is easily verified, this function plots as a strictly convex curve. Inasmuch as the second derivative at x - 7 is zero (rather than positive), this example serves to illustrate our earlier statement regarding the second derivative and the curvature of a curv^ (Sec, 9.3) to the effect that, while a positive f"(x) for all x does imply a strictly convex f (x), a strictly convex f (x) does not imply a positive f "{x) for all x. More importantly it also serves to illustrate the fact that, given a strictly convex (strictly concave) curve, the extremum found on that curve must be a minimum (maximum), because such ah extremum will either satisfy the second-order sufficient condition, or, failing that, satisfy another (higher-order) sufficient condition for a minimum (maximum). EXERCISE 9.6 1. Find the stationary values of the following functions: Determine by the Nth-derivative test whether they represent relative maxima, relative minima, or inflection points. 2. Find the stationary values of the following functions: (a) y = (x - 1 )3 + 16 (c) y = (3-x}6 + 7 (¿)y = (x-2)4 (d)y = < 5-2x)<+8 Use the Nth-derivative test to determine the exact nature of these stationary values. Chapter |

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