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Infinitesimals, a Relationship That Stretches Back Thousands of Years

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Infinitesimals, a Relationship That Stretches Back Thousands of Years J.M. HENLE Non-nonstandard Analysis: Real Infinitcsimals ~n athematics has had a troubled relationship with infinitesimals, a relationship that stretches back thousands of years. On the one ,hand, infinitesimals make intuitive sense. They're easy to deal with algebraically. They make calculus a lot of fun. On the other hand, they seem impossible to nail down. They're hard to deal "standard." Nor is it "nonstandard," however, as this term with intellectually. They can mask a fundamental lack of now has a well-def'med meaning. In what follows, we ma- understanding of analysis. nipulate sequences of real numbers. We treat them (mostly) There was hope, when Abraham Robinson developed as numbers. We add them, subtract them, and put them nonstandard analysis [R], that intuition and rigor had at last into functions. They aren't numbers, however. Trichotomy joined hands. His work indeed gave infinitesimals a foun- fails, for example. dation as members of the set of hyperreai numbers. But it The central construction in this article is a rediscovery. was an awkward foundation, dependent on the Axiom of Its first discoverer probably was D. Laugwitz. More on this Choice. Unlike standard number systems, there is no later. canonical set of hyperreal numbers. There is a way, however, of constructing infinitesimals Notation. We denote a sequence {an}nE ~ by a boldface naturally. Ironically, the seeds can be found in any calcu- a. We permit as to be undefined for fmitely many n. The lus book of sufficient age. At the turn of the century, it was key idea throughout is that of "from some point on." An typical of texts to define an infinitesimal as a "variable equation or inequality involving sequences will be in- whose limit is zero" [C]. That is the inspiration for the terpreted as being true or false, depending on whether present approach to calculus. Its infmitesimals are se- the associated equation or inequality involving the terms quences tending to 0. of the sequences is true or false from some point on. I call the system "non-nonstandard analysis" to draw at- For example, tention to its misfit nature. Having infinitesimals, it is not a>2 @ 1999 SPRINGERWERLAG NEW YORK, VOLUME 21, NUMBER 1, 1999 67 simply means "an > 2 from some point on," that is, that Proposition 1. Suppose a, b ~ 0. Then for some k, n -> k implies that an > 2. We use ordinary (1) a + b ~ 0. letters for real numbers. Note that a real number r can (2) a - b ~ 0. be viewed as the sequence a, with an = r for all n. (3) If e is finite, then ac ~ 0. (4) If Icl < lal, then e ~ O. Equations and inequalities involving sequences are inter- preted in the same way; that is, a + b = c means an § bn = PROOF. Given any positive d, we know that lal < d/2 and Cn from some point on, and sin(a) = b means sin(an) = bn Ibt < d/2. Then, la + b I -< lal + Ibl < d. Thisproves Part (1). from some point on. Note that forf(a) to be defined, it is Part (2) is proved similarly. only necessary that f(an) be defined from some point on. For Part (3), because Icl < r for some real r, and lal < The sequence b: 0, 0, 0, 1, 2, 3, 4,..., for example, does d/r for all positive d, we have lacl < d. have a reciprocal, I/b, because, as noted, finitely many For Part (4), observe that because lel < lal < d, Icl terms of a sequence may be undefined. <d. 9 If a statement P is true from some point on and state- ment Q is also true from some point on, then "P and Q" is Note that reals are finite, so that any result concerning fi- true from some point on. This fact allows us to do algebra nite sequences applies to reals too. Part (3) of Proposition l, on sequences. for example, tells us that if a ~ 0, then ra ~ 0 for any real r. Suppose, for example, we have a+4=3b Definition. For sequences a and b, we say that a and b are infinitely close, or a ~ b, iff a - b ~ 0. and c <45. Proposition 2. If a ~- r and b ~- s, then (1) a+b~r+s. Then, (2) a- b ~ r- s. (3) ab ~ rs. an + 4 = 3bn, from some point on, (4) If a -< s, then r -< s. and (5) a + b ~ r/s, if s r O. Cn < 45, from some point on. PROOF. Parts (1) and (2) follow easily from Proposition 1. Because the two are true together from some point on, For Part (3), note that (a-r)s, (b-s)r, and we can add them to get (a - r)(b - s) are all ~ 0. Adding these gives us ab - rs ~ O. an + 4 + Cn < 3bn + 45, from some point on. For Part (4), if r > s, then we would have both an -< s from some pont on and lan- r I < (r- s)/2 from some In other words, point on, which is impossible. a+4+c<3b+45. In view of Part (3), we need only show lfo ~ 1/s to establish Part (5). From s ~ b, we get Is~21 ~ Ib/21 < Ibl, so What we are doing, with this construction, is taking the by Part (4), Is~21 -< Ibl. Then I1/b - 1/s I = I(s - b)/bs I -< equivalence classes of sequences under the relation "equal- (1/Is/21)(1/Isb]s- b I. By Proposition 1, this is infinitesi- ity from some point on." To present it in that way, how- mal. 9 ever, would entail excessive formalism. One of my pur- This is all we need to get started. poses is to demonstrate that there is a fairly simple approach to infmitesimals, one that can reasonably be pre- Elementary Calculus sented to ordinary calculus students. Definition 1. A function f is continuous at x = r, r real, Infinitely Small and Infinitely Close iff Definition. A sequence a is infinitely small if lal < d a~rimpliesf(a)~f(r). for all positive real numbers d. For infinitely small a, we write a ~ 0. If lal < r for some real r, we say a is finite Equivalently, f is continuous at r if Ax ~ 0 implies or bounded. A sequence a is infinitesimal if a r 0 and f(r + Ax) -f(r) ~ O. a~0. We can also discuss continuity at a sequence a, as op- These definitions hide quantifiers. When we say that a r posed to a real r, but continuity on a set X corresponds to 0, we are actually saying that there is a k such that an r 0 continuity at all real numbers in X. As in nonstandard analy- for all n --> k. Similarly, this defmition of a ~ 0 is, in real- sis, continuity at all points (real numbers and sequences) ity, the more familiar and complicated Ve > 0 3k, n ~ k in a set X is equivalent to uniform continuity. I will prove lanl < e. this later. (}8 THE MATHEMATICAL INTELLIGENCER Proposition 3. The sum, difference, product, and quo- Definition 3. a is a subsequence of b (written a C b) if tient (when the divisor is nonzero) of functions contin- every term of a is a term of b; more precisely, if there is uous at x = r are continuous at x = r. an increasing function, k : N ~ N such that a n = bk(n). This follows easily from Proposition 2. Proposition 6. Let a C b be given. (1) Ifb is infinitesimal, then so is a. (2) If b ~ r, then a ~ r. Definition 2. For a functionfand a real r, we sayf'(r) = d iff for all infinitesimal Ax, (3) Ifb satisfies a given equation or inequality, then so does a. f(r + Ax) -f(r) ~ d. Ax The proof of this is routine. Here is what the computation looks like of the standard Proposition 7. (The Chain Rule). If y =fix) and z = first example: f(x) = x 2. g(y), then dz/dx = (dz/dy) 9(dy/dx). (x + Ax) 2 - x 2 x 2 - 2xAx + Ax 2 - x 2 Ax Ax PROOF. For Ax infinitesimal, dy/dx ~ Ay/Ax, where Ay = 2xAx + AX 2 f(x + Ax)-fix). By Proposition 4, Ay ~0, so seemingly Ax Az/Ay --~ dz/dy, where Az = g(y + Ay) -- g(y). But there is = 2x + Ax another way to write Az, since g(y + Ay) - g(y) = g(f(x) + ~2X. Ay) - g(f(x)) = g(f(x + Ax)) - g(f(x)), so dz/dx ~ Az/h,x. Proposition 4. If f is differentiable at r, it is continuous Putting this together, at r. dz Az Az Ay ~ dz dy PROOF. Just multiply dx Ax Ay Ax dy dx" The only difficulty with this is that Ay, while infinitely close f(r + Ax) - f(r) ~ d and Ax --~ O to 0, may not be infinitesimal because it may equal 0 infi- Ax nitely often. In that case, we can't claim that dz/dy .~ to get Az/Ay, because the definition of derivative requires Aye 0. f(r + Ax) -f(r) ~ 0. 9 But then, let Ax* C Ax be the subsequence such that The proofs of the differentiation rules are simple. The the corresponding Ay* is a sequence entirely composed of product rule is typical: O's. Then, dy/dx ~ Ay*/Ax* = 0. And, because Ay* = 0, the corresponding Az* = g(y + Ay*)- g(y) is also 0, so Proposition 5.
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