# 1.8无穷小病态三百年

1.8无穷小病态三百年

荷兰千年名校格罗宁根大学知名数学J.Ponstein教授为不的熟悉数理逻辑的数学工作者精心撰写了一部介绍“非标准分析”科普专著，对于我国高校微积分教学改革很有帮助。<span style=";color: white;font-family: "&font-size: 24px">

为此，我们将这部著作分为章节发表，加上适当小标题与评论。请读者注意。

袁萌 陈启清 9月29日

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1.8 Inﬁnitesimals in the 17th to the 19th century

There can be no doubt that in the 1670’s, some 1900 years after Archimedes lived, inﬁnitesimals were conceived by Leibniz. Moreover, he formulated their main properties, and many contemporary mathematicians as well as mathematicians after him, among them Euler and Cauchy, were able to successfully work with them. But the theory of the inﬁnitesimals lacked a rigorous basis, and during some 200 years all trials to improve this situation were in vein, so that at last one gave up, the more so because in the 1870’s Weierstrass came up with a rigorous theory of limits and continuity, which became the basis of what now is known as classical analysis, and where there was and is no need to consider inﬁnitesimals any more.

It is quite interesting to see how Euler [2] shows the well-known product formula for the sine function. He begins his proof with the equality, 2·sinh x = (1 + x/n)n −(1−x/n)n, valid for – in Eulers’s own words – ‘inﬁnitely large values’ of n. Obviously, this is only true up to an inﬁnitesimal. Then the right-hand side is treated as if n were

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a classical natural number. This leads after a purely classical reasoning to,

(1 + x/n)n −(1−x/n)n = (8x/n)·

m Y k=1

sin2(kπ/n)·{1 + x2/n2 tan(kπ/n)}, where m = (n−1)/2, taking n odd (the details of the reasoning do not matter here, and the case for n even is similar). So,

sinh x = (4x/n)·

m Y k=1

sin2(kπ/n)·{1 + x2/n2 tan2(kπ/n)}. Taking x 6= 0, and dividing by x, and then taking x = 0, gives, 1 = (4/n)· m Y k=1 sin2(kπ/n), and hence,

sinh x = x·

m Y k=1{1 + x2/n2 tan2(kπ/n)}. Now for k ﬁnite, n2 tan2(kπ/n) is ‘inﬁnitely close’ to (kπ)2, so (?)

sinh x = x·

∞Y k=1{1 + x2/k2π2}, and putting x = iz, this gives the desired result,

sinz = z·

∞Y k=1{1−z2/k2π2}. Obviously, at the question mark the argument goes a little too fast, and a number of steps must be included here (see e.g. Luxemburg [3]).

Another famous example is Cauchy’s proof ([4], p. 131), that a convergent series of continuous functions has a continuous limit function. To many this theorem was not correct, because it would seem that all kinds of counter-examples could be given. One of them is the series with the partial sums,

sn(x) = (4/π)·

n X k=1

sin(2k + 1)x 2k + 1

,

that is periodic modulo 2π and converges to, f(x) = −1 if −π < x < 0 0 if x = 0 or x = π +1 if 0 < x < π

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as can be shown by classical Fourier analysis. Since the sine function is everywhere continuous and sn(x) converges to f(x) for n tending to∞, according to Cauchy’s theorem f ought to be continuous, which it isn’t. But sofar, everything takes place within IR, and Cauchy let everything happen in what we have indicated by∗IR.

For him continuity of f at c meant that,∀x∈∗IR, x ' c : f(x) ' f(c), where, however, f :∗IR→∗IR and f need not be a standard function, and c∈∗IR, not only c∈IR, which is why his continuity is not∗continuity (in nonstandard analysis it is called S-continuity; recall deﬁnition (1.1) in Section 1.4, where c∈IR and a standard function was involved, so that there S-continuity was the same as∗continuity).

And by convergence of sn(x) to f(c) he meant that,∀n∼∞: sn(c) ' f(c), where again everything is in∗IR. Note that the Weierstrassian deﬁnitions of limit and continuity appeared half a century after Cauchy’s book, so Cauchy in a sense ‘had to’ work with deﬁnitions of the kind given here.

Now, by transfer,

∗sn(x) = (4/π)·

n X k=1

∗sin(2k + 1)x 2k + 1

,n∈∗IN, x∈∗IR,

and

∗f(x) = −1, or 0, or + 1, x∈∗IR, since if the range of a classical function f is ﬁnite, the range of its transform is the same as that of f. Let m∼ ∞be ﬁxed, and let x = c = 1/(2m), and dt = 1/m, so that x∼0, dt∼0. Then,∗sm1 2m= (2/π)· m X k=1∗sin(2k + 1)dt/2 (2k + 1)dt/2·dt. If we had that m∈IN, then the sum to the right would be an approximation of the Riemann-integral, J =Z1 0 sint t·dt,

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and it should therefore not come as a surprise that it can be shown that the standard part of the right-hand side is exactly equal to 2J/π, and hence,

∗sm1 2m−2J/π ' 0. But by direct calculation it follows that 2J/π 6= −1, 0, and +1, and since in particular for c = 1/(2m),∗f(c) = −1, or 0, or +1 (−1 is in fact impossible), it follows that∗sn(c) does not converge to∗f(c). Also, since for all n,∗sn(0) = 0,∗sm(x) is not continuous at c = 0, so that the ‘counter-example’ does not satisfy the assumptions of Cauchy’s theorem, and this is why Cauchy maintained his theorem against all criticism, but without basing his proof (and much of his other work) on a rigorous theory of the inﬁnitesimals and other nonstandard numbers. For many interesting details, see Lakatos [5].

更新时间： 2019-09-29 08:27:26

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