N n fXo f Xq t Xo2

This differs from the Maclaurin series of (9.8) only in the replacement of zero by x0 as the point of expansion and in the replacement of a by the expression (a — x0). What (9.13) tells us is that, given an «th-degree polynomial/(a), if we let a = 7 (say) in the terms on the right of (9.13). select an arbitrary number x0, then evaluate and add these terms, we will end up exactly with /(7)—the value of fix) at a = 7.

Example 2 Taking a0 = 3 as the point of expansion, we can rewrite (9.6) equivalently as f(x) = f(3) + /'(3)(a - 3) + - 3)2 + ■ • • + - 3)"

Expansion of an Arbitrary Function

Heretofore, we have shown how an nth-degree polynomial function can be expressed in another «th-degree polynomial form. As it turns out, it is also possible to express any arbitrary function 4>(x)—one that is not even necessarily a polynomial—in a polynomial form similar to (9.13), provided 4>(x) has finite, continuous derivatives up to the desired order at the expansion point x0.

According to a mathematical proposition known as Taylor's theorem, given an arbitrary function <j>(x), if we know the value of the function at x = x0 [that is. <f>(-*0)] and the values of its derivatives at .*„ [that is. <fi'(x0). 4>"(x0), etc.], then this function can be expanded around the point a0 as follows (n = a fixed positive integer arbitrarily chosen):

R, where Pn represents the (bracketed) «th-degree polynomial [the first (n 4- 1) terms on the right], and Rn denotes a remainder, to be explained below.* The presence of Rn distinguishes (9.14) from (9.13), and for this reason (9.14) is called a Taylor series with remainder. The form of the polynomial Pn and the size of the remainder Rn will depend on the value of n we choose. The larger the n. the more terms there will be in Pn\ accordingly, Rn will in general assume a different value for each different n. This fact explains the need for the subscript n in these two symbols. As a memory aid, we can identify n as the order of the highest derivative in Pn. (In the special case of n = 0, no derivative will appear in Pn at all.)

The appearance of Rn in (9.14) is due to the fact that we are here dealing with an arbitrary function $ which cannot always be transformed exactly into the polynomial form shown in (9.13). Therefore, a remainder term is included as a supplement to the Pn part, in order to represent the difference between <p(x) and the polynomial Pn. Looked at differently, Pn may be considered a polynomial approximation to <t>(x), with the term Rn as a measure of the error of approximation. If we choose n = 1, for example, we have

4>{x) = [0(x0) + <P'(x0)(x - x0)] + «,=/>, + Rt where P\ consists of n + 1 = 2 terms and constitutes a linear approximation to <i>(x). If we choose n = 2, a second-power term will appear, so that

Ri where P2, consisting of n + 1 = 3 terms, will be a quadratic approximation to <p(x). And so forth.

* The symbol Rn (remainder) is not to be confused with the symbol R" (n-space).

We should mention, in passing, that the arbitrary function <j>(x) could obviously encompass the nth-degree polynomial of (9.6) as a special case. For this latter case, if the expansion is into another nth-degree polynomial, the result of

(9.13) will exactly apply; or in other words, we can use the result in (9.14), with Rn = 0. However, if the given nth-degree polynomial /(x) is to be expanded into a polynomial of a lesser degree, then the latter can only be considered an approximation to /(x), and a remainder will appear; accordingly, the result in

(9.14) can be applied with a nonzero remainder. Thus the Taylor series in the form of (9.14) is perfectly general.

Example 3 Expand the nonpolynomial function around the point x0 = 1, with n = 4. We shall need the first four derivatives of 4>(x), which are

<j>'(x) = -(1 + x)-2 so that <f>'(l) = ~(2y2 = ~ 1

0 0

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