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What do people mean when they say that Euler treated infinity differently? I read in various books that, today, mathematicians would not approve of Euler's methods and his proofs lacked rigor. Can anyone elaborate?

Edit: If I remember correctly Euler's original solution to the Basel problem is as follows.

Using Taylor series for $\sin (s)/s$ we write $$\sin (s)/s = 1 - {s^2}/3! + {s^4}/5! - \cdots $$ but $\sin (s)/s$ vanishes at $\pm \pi$, $\pm 2\pi$, etc. hence $$\frac{{\sin s}}{s} = {\left( {1 - \frac{s}{\pi }} \right)}{\left( {1 + \frac{s}{\pi }} \right)}{\left( {1 - \frac{s}{{2\pi }}} \right)}{\left( {1 + \frac{s}{{2\pi }}} \right)}{\left( {1 - \frac{s}{{3\pi }}} \right)}{\left( {1 + \frac{s}{{3\pi }}} \right)} \cdots$$ or $$\frac{{\sin s}}{s} = {\left( {1 - \frac{{{s^2}}}{{{1^2}\pi^2}}} \right)}{\left( {1 - \frac{{{s^2}}}{{{2^2}{\pi ^2}}}} \right)}{\left( {1 - \frac{{{s^2}}}{{{3^2}{\pi ^2}}}} \right)} \cdots$$ which is $$\frac{{\sin s}}{s} = 1 - \frac{{{s^2}}}{{{\pi ^2}}}{\left( {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \cdots } \right)} + \cdots.$$ Equating coefficients yields $$\zeta (2) = \frac{{{\pi ^2}}}{6}.$$

But $\pm \pi$, $\pm 2\pi$, etc. are also roots of ${e^s}\sin (s)/s$, correct? So equating coefficients does not give ${\pi ^2}/6$.

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Dear glebovg, Euler was surely aware that $(e^s \sin s)/s$ had the same zeroes of $(\sin s)/s$, but presumably also understood that the former had a much faster rate of growth than the latter, and so didn't have the same "polynomial-like" behaviour. Regards, –  Matt E Oct 22 '12 at 3:04
    
But we can still factor ${e^s}\sin (s)/s$ just like we factored $\sin (s)/s$. –  glebovg Oct 22 '12 at 3:16
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Dear glebovg, I'm not sure what you mean, but let me be more precise: for two polynomials $f$ and $g$, we know that if $f$ and $g$ have the same zeroes, and the ratio $f(x)/g(x)$ goes to $1$ as $x \to \infty$, then $f = g$. (Checking the limit at infinity removes the ambiguity scaling.) Similary, Euler's infinite product for $(\sin s)/s$, and the function $(\sin s)/s$ itself, have the same zeroes, and their ratio is well-behaved as $s \to \infty$. This is not true if we divide $e^s (\sin s)/s$ by the infinite product: while we get a zero-free function, it blows up at infinity. Regards, –  Matt E Oct 22 '12 at 3:36
    
P.S. I should say that of course, ultimately we know that the ratio of $(\sin s)/s$ and the infinite product is equal to $1$, but I am claiming we can prove this by making an analysis of the growth of the ratio, and my guess is that Euler would have been sensitive to this sort of phenomenon and possibility: he was, after all, one of the very greatest mathematicians of all time. Regards, –  Matt E Oct 22 '12 at 3:39
    
@glebovg, Note that Euler did not jump from the equality of zeros to the equality of sine to the infinite product. He gave an elaborate, and essentially correct, argument in favor of such equality. –  user72694 Apr 24 at 17:35

4 Answers 4

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Basically they just mean that many of his arguments involving, for example, infinite products and sums are not rigorous by modern standards. Sometimes, for instance, he manipulated them in ways that make sense for finite products and sums but that we now know don’t always make sense for infinite products and sums. Fortunately, he was an extraordinarily good mathematician and had an excellent sense of when these manipulations would actually work.

In particular, Euler predates rigorous notions of convergence, so his proofs ignore convergence issues. An example can be seen in this sketch of his proof of the product formula for the zeta function: he simply carried out the infinite manipulations, but by modern standards of rigor they require some justification.

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This answer would have some validity had Euler indeed limited himself to pointing out that sine and the infinite product have the same zeros, and jumped to the conclusion that they must be equal. This however is not what Euler did. Instead he provided an elaborate 7-step argument in favor of the decomposition. In fact his argument can be formalized, step-by-rigorous-step, in the framework of modern theories, as discussed in this recent article. –  user72694 Apr 24 at 17:34

There is another way Euler "lacked rigour" in nowadays terms.

He used the idea of "something infinitesimally small" in his Introductio in analysin infinitorum (chapter 7, §115). He just gave this meaning to a variable and identified the term with its limit. So he would have said "$\frac{1}{\delta}=0$ for $\delta$ infinitely small". (This is something people use to do nowadays - at least when they aren't mathematicians.)

Clearly Euler didn't have the notions of mathematics from Cauchy, Weierstrass and so on. So it's kind of mean to say he lacked rigour. (By the way: I recommend reading (or at least browsing) the Introductio once - it is quite interesting to see how he develops all these equalities between trigonomic, rational and exponential functions.)

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So, in some sense, he disregarded the Archimedean property of the real numbers. –  glebovg Oct 18 '12 at 16:13
    
I don't think he went that far. It was just his way of explaining this intuition. He didn't thinkt about its (algebraic-)theoretical implications. Besides I'm not sure if this principle was known in his time. –  AndreasS Oct 18 '12 at 16:33
    
I think he must have realized that if $\delta $ is infinitesimal then multiplying it by $1/\delta $ gives 1. Perhaps he believed in the existence of infinitesimals because it makes sense in philosophy. –  glebovg Oct 18 '12 at 16:39
    
I believe George Berkeley wrote a book about infinitesimals and Newton's reasoning, claiming that it was nonsense. –  glebovg Oct 18 '12 at 16:46
    
Newton's reasoning weren't the most concise - as far as I know. George Berkley argued against infinitesimals but he argued against mathematics as a whole, as well. I'd say Euler just used this way as a "façon de parler" but could not make satisfactory sense of it - as nobody could at that time. Besides, the Introducio was a textbook. Maybe Euler simplified his reasoning there a little (although he certainly didn't think in "$\epsilon$-terms") –  AndreasS Oct 18 '12 at 17:25

Euler worked before calculus was placed on rigorous foundations by Cauchy, Riemann and Weierstrass. One of his favorite techniques was to exploit analogies between polynomials and power series, viewing power series as polynomials of infinite degree. His keen intuition allowed him to avoid pitfalls, often obtaining results that could be later translated into rigorous proofs. Below is a prototypical example, excerpted from historian Judith V. Grabiner's Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus.

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Convergent power series can indeed be manipulated like polynomials of infinite (more precisely, hyperfinite) degree. This is the point that Grabiner and a number of other historians missed. The record has been set straight by Laugwitz in a series of papers in top history journals; see e.g., here, and other historians have followed his lead, resulting in numerous publications over the past two decades that have started the set the record straight on Euler. –  user72694 Apr 25 at 8:16
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@use I'm familiar with the work of Laugwitz etc. However, even though one can devise rigorous interpretations of some of Euler's methods using later research, that does not necessarily imply that those were Euler's methods, or that Euler's methods were rigorous. In matters like this, if one strives to write an accurate history, one needs to be very careful when assessing such claims because it is far too easy to mistakenly inject modern ideas where they never existed. It takes tremendous effort to strive to understand the knowledge that existed in certain periods, schools and individuals. –  Bill Dubuque Apr 25 at 13:09
    
Bill, I agree with you fully. These are the issues that are insufficiently well understood and that we are currently trying to understand. We have a current text on Euler that I think may answer your question to your satisfaction. This is in advanced stages of refereeing at a leading philosophy journal. If you are interested please contact me via email so I can send you a copy of the current version. –  user72694 Apr 26 at 18:47

To respond to your question concerning Euler's treatment of infinity, note that Euler did indeed "treat infinity differently" from the way it is viewed today. Our conceptual framework today is dominated by the work of Cantor, Dedekind, and Weierstrass, who sought to eliminate infinitesimals and replace them by epsilon, delta procedures within the context of an Archimedean continuum devoid of infinitesimals. Euler, on the other hand, worked with infinitesimals galore, and used infinite numbers freely. Thus he viewed an infinite series as a polynomial of infinite order. In the terminology of the historian Detlef Laugwitz, his arguments contained some "hidden lemmas" that require further justification, which can indeed be provided in light of modern theories.

Other than that, Euler's techniques and procedural moves are closely mirrored by techniques and principles developed in the context of a hyperreal extension $\mathbb{R}\subset{}^{\ast}\mathbb{R}$, and his "infinite numbers" admit of proxies in the hyperreal approach, namely hyperreal integers in $^\ast\mathbb{N}\setminus\mathbb{N}$. Thus, an infinite series is approximated (up to infinitesimal error) by a polynomial of infinite hyperfinite degree. These can be manipulated like ordinary polynomials by the transfer principle.

Euler obviously did not have the semantic foundational frameworks developed from 1870 onward such as ZFC, but his syntactic procedures are successfully and faithfully mirrored in the hyperreal approach. Historians critical of Euler's techniques are generally ignorant of hyperreal techniques and therefore hostile toward them.

A number of articles in the literature successfully interpret Euler's procedures in terms of modern infinitesimals (with the syntactic/semantic proviso stated above), including the work of Kanovei, Laugwitz, McKinzie, Tuckey, and others.

Note that Euler did not jump from the equality of zeros to the equality of sine to the infinite product. He gave an elaborate, and essentially correct, argument in favor of such equality. More specifically, Euler provided an elaborate 7-step argument in favor of the decomposition. In fact his argument can be formalized, step-by-rigorous-step, in the framework of modern theories, as discussed in this recent article.

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I completely agree that historians tend to ignore Robinson's hyperreals. I think they are ignorant of nonstandard analysis because nonstandard analysis relies on the axioms and theorems of set theory. –  glebovg Oct 20 '13 at 23:37
    
@glebovg, thanks for your comment. I think the situation is more complicated than that. Traditional epsilontic analysis similarly relies on axioms and theorems of set theory. Rather than speculating as to the reasons for such an attitude on the part of historians, it may be more fruitful to try to establish a more balanced historical perspective on some key figures such as Cauchy; see e.g. a current thread here. –  user72694 Oct 21 '13 at 13:41

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