# How did Gauss discover the invariant density for the Gauss map?

The Gauss map is defined on $(0,1)$ by the formula $$f(x)=\frac1x-\Big\lfloor\frac1x\Big\rfloor$$ Then the density $$\rho(x)=\frac{1}{\log2(1+x)}$$ is $f$-invariant.

It appeared in Gauss' diary. Gauss didn't indicate the way he had found the density. Checking invariance is straightforward.

Is there a simple (short) way to come up with this density function?

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A. Ya. Khinchin says in the little book "Continued Fractions" that Gauss did not include a proof in the letter to Laplace that gives the density, never published the proof, but these appeared with Kuz'min in 1928 and Levy in 1929. Khinchin indicates that his proof follows Kuz'min. But it is several pages. Available as a Dover paperback. And, of course, there could be a modern short proof. –  Will Jagy May 3 '10 at 1:40
Will, I think Khinchin is talking about the proof of ergodicity of $f$. This is important as it can be used to proof a statement on distribution of continued fraction "digits". This statement is frequently referred to as "Kuz'min's theorem". I may be wrong –  Zarathustra May 4 '10 at 1:12
Yes, Khinchin is preparing the whole topic in very few pages, the Gauss measure blends in with lots of related stuff. Section 15, "Gauss's problem and Kuz'min's theorem" starts on page 71, and I think the first time he mentions the exact form of your $\rho(x)$ is formula (80) on page 82. I see you liked Peter Luthy's heuristic. Good. –  Will Jagy May 4 '10 at 1:56

Here is a heuristic.

Observe that $f$ is decreasing on any interval of the form $I_n:=(1/(n+1),1/n)$. If $x\in I_n$ and $f(x)=a$, then $x=\frac{1}{a+n}$, and so $f^{-1}(a,1)$ is the union of the intervals $(1/(n+1),1/(n+a))$. Supposing there were an $f$-invariant measure $\mu=gdx$, you can see that

$\displaystyle{\sum_{n=1}^\infty \int_{1/n+1}^{1/n+a}g(x)dx=\int_a^1g(x)dx}$.

Supposing that $g$ is continuous, take the derivative of both sides:

(**) $\displaystyle{\sum_{n=1}^\infty g\left(1/(n+a)\right)/(n+a)^2=g(a)}$.

Approximating the sum by an integral, we get

$\displaystyle{g(a)\approx\int_1^\infty g\left(\frac{1}{x+a}\right)\frac{1}{(x+a)^2}dx}$.

Supposing that $g$ is never too big or too small, this gives

$g(a)\approx \frac{C}{1+a}$.

One can then plug this kind of function into (**) to see that such a function works for any C. Then just pick C to normalize.

Probably someone from the era of Gauss (especially with his acuity at mathematics) did not need to do anything past (**) since people back then seem like magicians when it comes to expressions with infinite sums.

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This is nice! Thank you! –  Zarathustra May 4 '10 at 1:13
You're very welcome! –  Peter Luthy May 4 '10 at 5:47
It is not the density. It is distribution function. Density function $1/(x+1)$ is not so complicated to find it. Rational numbers lead to this function as well. So some experiments can give this function.
You can consider continued fraction expansion for rational numbers $a/b$, $1\le a\le b\le R$, $R\to\infty$. You will find that there is probability $p(1)$ that partial quotient is equal to 1 (or 2, 3,...). More generally there ia probability that the tail of continued fraction is less or equal to the fixed number $x$. Using $p(1)$, $p(2)$... you'll find values of distribution function at $x=1/2$, $2/3$, $3/4$,... Next spep is to find appropriate density function. –  Alexey Ustinov May 3 '10 at 7:10