Has anyone published in print or on a web site or elsewhere a compilation of successful illogical formal arguments? By those I mean arguments that follow a form in disregard of the legality of its applicability in the circumstances to which it is applied, and get a right answer.

One notable example occurred when a formula was discovered for finding solutions of third-degree algebraic equations with real coefficients, when it was known that a real root exists. It was necessary to take square roots of negative numbers. These "imaginary" numbers canceled out and then (the punch line): when the resulting values were substituted into the equation, it checked.

Many things that Euler did could be considered examples. It has been noted here in stackexchange that Euler's derivation of the product formula for the sine function considered the infinitely large leading coefficient of the polyonomial that was the product (which is an infinite product and so has no leading coefficient).

Paul Dirac's delta function was seen to lead to demonstrably correct results at a time when it was not known to to have any logically rigorous justification.

Here's an example from Paul Halmos' article Does Mathematics Have Elements:

In the general ring theory question there are no numbers, no absolute values, no inequalities, and no limits - those concepts are totally inappropriate and cannot be brought to bear. Nevertheless an impressive-sounding classical phrase, "the principle of permanence of functional form", comes to the rescue and yields an analytically inspired proof in pure algebra. The idea is to pretend that $1/(1-ba)$ can be expanded in a geometric series (which is utter nonsense), so that $$(1-ba)^{-1} = 1 + ba + baba + bababa + \cdots.$$ It follows (it doesn't really, but it's fun to keep pretending) that $$(1-ba)^{-1} = 1 + b(1 + ab + abab + ababab + \cdots )a,$$ and, after one more application of the geometric series pretense, this yields $$(1-ba)^{-1} = 1 + b(1-ab)^{-1}a.$$ Now stop the pretense and verify that, despite its unlawful derivation, the formula works. If, that is, $c = (1-ab)^{-1}$, so that $(1-ab)c = c(1-ab) = 1$, then $1 + bca$ is the inverse of $1 - ba$. Once the statement is put this way, its proof becomes a matter of (perfectly legal) mechanical computation. Why does it all this work? What goes on here? Why does it seem that the formula for the sum of an infinite geometric series is true even for an abstract ring in which convergence is meaningless? What general truth does the formula embody? I don't know the answer, but I note that the formula is applicable in other situations where it ought not to be,[ . . . ]

Question: Has anyone published in print or on a web site or elsewhere a compilation of examples of this phenomenon? (Maybe annotated with explanations of the resolution of the seeming paradox in cases where it is known.)

  • 7
    $\begingroup$ Umbral calculus is another example of this. Also 17th-century use of infinitesimals, which weren't put on a rigorous footing for 300 years. $\endgroup$ – MJD Jul 7 '12 at 23:42
  • 1
    $\begingroup$ There are various ways to "explain" why the "proof" quoted by Halmos works - see my MO question: How would you solve this tantalizing Halmos problem? $\endgroup$ – Bill Dubuque Jul 7 '12 at 23:52
  • 1
    $\begingroup$ There are also some arguments about regular languages from infinite series. The Kleene closure $L^*$ of a language $L$ is defined to be $\bigcup_{i=0}^\infty L^i$, or $1 + L + LL + LLL +\ldots$ where $1$ is the language containing only the empty string, $+$ is set union and (invisible) multiplication is regular language concatenation. Then you can say $L^* = (1-L)^{-1}$ as if this makes sense, although subtraction of languages is risible. I forget how it goes from there, but you can actually get a useful conclusion out of this nonsense, so there must be something to it. $\endgroup$ – MJD Jul 7 '12 at 23:57
  • 8
    $\begingroup$ I think the MathOverflow thread on "jokes in the sense of Littlewood" is in the same spirit. $\endgroup$ – Rahul Jul 8 '12 at 0:57
  • 4
    $\begingroup$ Sounds like a good candidate for a big-list question... $\endgroup$ – Nate Eldredge Jul 8 '12 at 3:15

I think the MathOverflow thread on "jokes in the sense of Littlewood" is in the same spirit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.