Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Simple continued fraction
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Typical continued fractions=== Most irrational numbers do not have any periodic or regular behavior in their continued fraction expansion. Nevertheless, for [[almost all]] numbers on the unit interval, they have the same limit behavior. The arithmetic average diverges: <math>\lim_{n\to\infty}\frac 1n \sum_{k=1}^n a_k = +\infty</math>, and so the coefficients grow arbitrarily large: <math>\limsup_n a_n = +\infty</math>. In particular, this implies that almost all numbers are well-approximable, in the sense that<math display="block">\liminf_{n\to\infty} \left| x - \frac{p_n}{q_n} \right| q_n^2 = 0</math>[[Aleksandr Khinchin|Khinchin]] proved that the [[geometric mean]] of {{math|''a''<sub>''i''</sub>}} tends to a constant (known as [[Khinchin's constant]]):<math display="block">\lim_{n \rightarrow \infty } \left( a_1 a_2 ... a_n \right) ^{1/n} = K_0 = 2.6854520010\dots</math>[[Paul Lévy (mathematician)|Paul Lévy]] proved that the {{mvar|n}}th root of the denominator of the {{mvar|n}}th convergent converges to [[Lévy's constant]] <math display="block">\lim_{n \rightarrow \infty } q_n^{1/n} = e^{\pi^2/(12\ln2)} = 3.2758\ldots</math>[[Lochs' theorem]] states that the convergents converge exponentially at the rate of<math display="block">\lim_{n\to\infty}\frac 1n \ln\left| x - \frac{p_n}{q_n} \right| = -\frac{\pi^2}{6\ln 2} </math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)