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
Geometric mean
(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!
===Iterative means=== The geometric mean of a data set [[inequality of arithmetic and geometric means|is less than]] the data set's [[arithmetic mean]] unless all members of the data set are equal, in which case the geometric and arithmetic means are equal. This allows the definition of the [[arithmetic-geometric mean]], an intersection of the two which always lies in between. The geometric mean is also the '''arithmetic-harmonic mean''' in the sense that if two [[sequence]]s (<math display="inline">a_n</math>) and (<math display="inline">h_n</math>) are defined: :<math>a_{n+1} = \frac{a_n + h_n}{2}, \quad a_0 = x</math> and :<math>h_{n+1} = \frac{2{a_n}{h_n}}{a_n + h_n}, \quad h_0 = y</math> where <math display="inline">h_{n+1}</math> is the [[harmonic mean]] of the previous values of the two sequences, then <math display="inline">a_n</math> and <math display="inline">h_n</math> will converge to the geometric mean of <math display="inline">x</math> and <math display="inline">y</math>. The sequences converge to a common limit, and the geometric mean is preserved: :<math>\sqrt{a_{i+1} h_{i+1}} = \sqrt{\frac{a_i + h_i}{2}\frac{2{a_i}{h_i}}{a_i + h_i}} = \sqrt{{a_i}{h_i}} </math> Replacing the arithmetic and harmonic mean by a pair of [[generalized mean]]s of opposite, finite exponents yields the same result.
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)