Why does the continued fraction of Champerowne's number have so large entries?

Question

If we calculate the continued fraction of Champerowne's number $$0.1234567891011121314151617181920\cdots $$ it turns out that it contains very large entries

How can I understand this intuitively ?

The decimal representation of Champerowne's number must have parts which make the number look like a rational because of appearing periods, but it is not obvious where these (large) periods should be.

Many zeros (making the number look like to have a terminating decimal expansion) occur later, so they do not explain the large entries.

Any ideas ?

Answer 1

As a first-order approximation, it is very close to

$ 10/81 = 0.1 + 0.02 + 0.003 + \dots = 0.123456790123\dots $

As a second,

$ 1/9801 = 0.0001 + 0.000002 + \dots = 0.0001020304050607080910111213\dots $

so it should not come as a surprise that Champernowne's [sic] number is close to the sum of $10/81$ and another rational which is a multiple of $1/(9801\cdot10^8)$, the factor $10^8$ being needed to shift the $10111213\dots$ part 8 decimal places to the right, to leave the initial $0.12345678$ alone (the $9$ is already in the correct place w.r.t. the following $10$). In fact

$ 10/81-1002/980100000000 \approx 0.12345678910111213\dots96979900010203040506070809\dots$

Answer 2

Warning: I am leaving the answer below while I try to repair it, but I now believe it to be seriously incomplete, if not misleading. It is certainly true that if the continued fraction of a constant contains extremely large terms, then the corresponding convergents will be exceptionally good rational approximations of the constant. But a constant having exceptionally good rational approximations does not imply that there will be large terms in its continued fraction, so merely noting the existence of these exceptionally good rational approximations is not a sufficient explanation for why the continued fraction has such large terms. A useful counterexample to keep in mind is the constant $$ \sum_{n=0}^\infty\frac{1}{10^{2^n}}. $$ The inclusion of each successive term in this series doubles the number of correct digits in the decimal expansion, so just as I say in my answer with regard to the Champernowne constant, "Each additional term in this series provides an exponentially growing number of correct digits...." Nevertheless, the continued fraction contains no large terms. In fact, the terms have a simple pattern and, more importantly, are bounded, as was shown in

Jeffrey Shallit, Simple continued fractions for some irrational numbers, Journal of Number Theory 11 (1979) 209-217.

Interestingly the Liouville constant and the constant $$ \sum_{n=0}^\infty\frac{1}{10^{3^n}}, $$ like the Champernowne constant, exhibit the large term phenomenon in their continued fractions. All three continued fractions contain a sequence of terms that grows super-exponentially. The continued fraction of the constant $$ \sum_{n=0}^\infty\frac{1}{10^{2^n-n}} $$ has large terms as well, but they grow only exponentially, not super-exponentially. That the terms in the continued fraction of $$ \sum_{n=0}^\infty\frac{1}{10^{2^n}} $$ do not grow at all indicates that the issues are rather subtle and that careful analysis is needed.

The original answer: To expand a bit on Rosie F's answer, we can write \begin{align} C_{10} =& \frac{10}{81} - \left(\frac{91}{81}-\frac{991}{9801}\right)\times10^{-9}-\left(\frac{9901}{9801}-\frac{99901}{998001}\right)\times10^{-189}\\ &-\left(\frac{999001}{998001}-\frac{9999001}{99980001}\right)\times10^{-2889}-\left(\frac{99990001}{99980001}-\frac{999990001}{9999800001}\right)\times10^{-38889}+\ldots. \end{align} One can see that the first term of this series approximates $C_{10}$ with an error of about $10^{-9}$, the sum of the first two terms approximates $C_{10}$ with an error of about $0.9\times10^{-189}$, the sum of the first three terms approximates $C_{10}$ with an error of about $0.9\times10^{-2889}$, and so on. Each additional term in this series provides an exponentially growing number of correct digits of $C_{10}$.

To see where the series comes from, write the base-$b$ Champernowne constant as a double sum, $$ C_b=\sum_{n=1}^\infty b^{-E_b(n-1)}\sum_{k=b^{n-1}}^{b^n-1}\frac{k}{b^{n(k-b^{n-1}+1)}}, $$ where $E_b(n)=\sum_{k=1}^nk(b-1)b^{k-1}$. Notice that this is just a direct translation of the definition of the constant: the inner sum runs over all the $n$-digit base-$b$ numbers, giving, for example, $$0.1+0.02+0.003+\ldots+0.000000009=0.123456789$$ when $b=10$ and $n=1$ and $$0.10+0.0011+0.000012+\ldots+0.\underbrace{00\ldots0}_{\text{178 zeroes}}99=0.101112\ldots9899$$ when $b=10$ and $n=2$; the pre-factor of this sum just shifts the sum to the appropriate starting position $E_b(n-1)$ places to the right of the radix point. (This is the number of places occupied by base-$b$ numbers of fewer than $n$ digits.)

Both the expression for $E_b(n)$ and the inner sum of the expression for $C_b$ are of the form $C\sum_{k=p}^q ka^k$, which can be evaluated using $$ \sum_{k=p}^q ka^k=a^p\frac{p-(p-1)a}{(1-a)^2}-a^{q+1}\frac{q+1-qa}{(1-a)^2}. $$ This can be used to write $$ E_b(n) = nb^n-\frac{b^n-1}{b-1} $$ and $$ b^{-E_b(n-1)} \sum_{k=b^{n-1}}^{b^n-1}\frac{k}{b^{n(k-b^{n-1}+1)}} = \frac{b^{2n-1}-b^{n-1}+1}{\left(b^n-1\right)^2}b^{-E_b(n-1)}-\frac{b^{2n}-b^n+1}{\left(b^n-1\right)^2}b^{-E_b(n)}, $$ from which we get $$ C_b = \frac{b}{(b-1)^2}-\sum_{n=1}^\infty\left(\frac{b^{2n}-b^n+1}{\left(b^n-1\right)^2} - \frac{b^{2n+1}-b^n+1}{\left(b^{n+1}-1\right)^2}\right)b^{-E_b(n)}. $$ This produces the series for $C_{10}$ at the beginning of this post.

The rational numbers one gets by truncating this series at $n=N$ for some finite $N$ generally agree with the convergents one gets by truncating the continued fraction of $C_b$ just before one of the high water marks. (A high water mark is a coefficient in the continued fraction larger than all previous coefficients.) There are some exceptions when $b$ and $N$ are both very small.

Remark on the intuition: I wanted to address this part of your post:

How can I understand this intuitively ?

The decimal representation of Champerowne's number must have parts which make the number look like a rational because of appearing periods, but it is not obvious where these (large) periods should be.

Intuitively you can say that a real number looks rational if its decimal expansion agrees for many places with that of a rational number whose numerator and denominator have far fewer digits than has the string of places on which the agreement occurs. For example, which of the following two numbers look rational? \begin{align} &0.01886792452830\ldots\\ &0.01818181818181\ldots \end{align} You can say that both look rational since they both agree to a lot of places with a rational number that has two-digit denominator, the first with $\frac{1}{53}$, the second with $\frac{1}{55}$. That the second appears to have a period of length $2$ does not make it closer to rational than the first. The Champernowne constant is similar to the first of these examples in that there's not going to be a short periodic pattern close to the decimal point to clue you in to the fact that the constant is very close to a rational number.