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My question is,

In the history of mathematics, what are some non-rigorous approaches that lead to a correct result?

One example I can think of is Euler’s initial approach to solve the Basel problem.

His argument is

Because $n\pi$ is a root of $\sin x$, so $$\sin x=x(1+\frac{x}\pi)(1-\frac{x}{\pi})(1+\frac{x}{2\pi})(1-\frac{x}{2\pi})\cdots$$

Equating the Taylor series and the product series of $\frac{\sin x}x$ and compare the coefficients of $x^2$, one obtains $$-\frac1{\pi^2}\sum^\infty_{n=0}\frac1{n^2}=-\frac16$$ which solves the problem.

Obviously, the expansion of function in terms of its roots is not always correct, and requires the justification of Weierstrass factorization theorem, which was not available at the time.

What are some other (famous) examples?

Szeto
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1 Answers1

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Surprisingly, there was no logical definition of $\Bbb R$ and no rigorous foundation for basic calculus, until the 19th century. In the 18th century some theologians mocked mathematicians for their belief in infinitesimals. A vast corpus of original discovery was achieved with what, by today's standards, was a non-rigorous base. Here is an example of a style of proof that was perfectly acceptable in the 17th century:

If $a_n\geq a_{n+1}\geq 0$ for each $n\in \Bbb N$ and if $\sum_{n=1}^{\infty}a_n$ converges then $\lim_{N\to \infty}Na_N=0.$ PROOF: For $M,N \in \Bbb N$ with $M<N$ we have $$0\leq Na_N=\frac {1}{1-\frac {M}{N}}(N-M)a_N\leq \frac {1}{1-\frac {M}{N}}\sum_{n=M+1}^Na_n.$$ Now if $M$ is infinitely large then $\sum_{n=M+1}^{\infty}a_n$ is infinitely small, and we have $0\leq\sum_{n=M+1}^Na_n\leq \sum_{n=M}^{\infty}a_n,$ so $\sum_{n=M+1}^N a_n$ is infinitely small. And when $N$ is infinitely larger than $M$ the value of $\frac {1}{1-\frac {M}{N}}$ is infinitely close to $1$.

  • Haha. The proof makes no sense, especially for the last line: what if I set $N=M+1$? – Szeto Jul 23 '18 at 02:55
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    You can't set $N=M+1$ because $N$ is infinitely larger than $M$. That is, $M/N$ is infinitely close to $0$. You can transform this into a modern "$\epsilon$-$\delta$" proof with no difficulty – DanielWainfleet Jul 23 '18 at 03:33