Geometric mean ($G=(x_1\cdots x_n{)}^{1/n}$)

The geometric mean replaces a list of nonnegative real numbers by one repeated value that preserves their product of numbers. For $n\ge 1$ inputs,For two positive inputs, $G=\sqrt{x_1{x}_2}$ is their positive square root. The geometric mean of $1$ and $9$ is three: replacing $1\cdot 9$ by $3\cdot 3$ retains the same product. In contrast, their arithmetic mean five retains the sum $1+9=5+5$. The geometric mean is zero if any input is zero; the logarithm formula below requires every input to be positive. ^[ref].

Its usefulness becomes immediate for successive growth factors. Suppose a hypothetical positive quantity starts at $Q_0=1000$, is multiplied by $3/2$, then by $2/3$. It goes $1000\to 1500\to 1000$, so the geometric mean factor is $\sqrt{(3/2)(2/3)}=1$: a constant factor of one for both steps gives the same endpoint. The arithmetic mean factor is $13/12$; repeating it would give $1000(13/12{)}^2\approx 1173.61$, a different endpoint. A mean of factors summarizes multiplicative change; subtracting one afterwards gives the equivalent change per step. Equal-duration steps are necessary to interpret this as change per common time interval.

For positive inputs, logarithms turn multiplication into addition, soThus the geometric mean is an arithmetic mean in logarithmic coordinates, followed by the exponential function. For positive measurements with a common unit, one can use dimensionless ratios to a fixed reference $x_0$: $G=x_0\exp (n^{-1}{\sum }_i\log (x_i/x_0))$. Changing the common unit rescales $G$ by the same factor as the inputs. As a check, doubling one input and halving another retains $G$ while usually changing the arithmetic mean.

The arithmetic-geometric mean inequality gives $G\le A$. Apply that same inequality to the multiplicative inverses $1/x_i$ to obtain $H\le G$, where $H$ is the harmonic mean. Consequentlyfor positive inputs, with equality throughout exactly when all inputs agree. For $30$ and $60$, the three values are $H=40$, $G=30\sqrt{2}\approx 42.4264$, and $A=45$. They preserve, respectively, the sum of inverses, the product, and the sum. No one of these summaries preserves all three for unequal inputs.

For positive quantities measured at successive steps, the growth factor $g_i=Q_i/Q_{i-1}$ is a dimensionless ratio. It can be below one when the quantity shrinks. Repeated multiplication gives $Q_n=Q_0{\prod }_{i=1}^n{g}_i$, so the common factor that reproduces the endpoint after $n$ identical steps is their geometric mean. A factor $g$ corresponds to a relative change $g-1$; factors must be averaged before subtracting one when summarizing compounded change.

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For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted geometric mean isIt takes a weighted mean of logarithms. A zero weight excludes that input; equal weights give the ordinary geometric mean. If $x_i$ occurs $m_i$ times in a list of total length $M$, its correct weight is $w_i=m_i/M$. For the list $(12,3,3,3)$,and replacing all four entries by $G_w$ preserves the product $324$.

There is also an optimization interpretation. A logarithmic transformation turns multiplicative discrepancies into additive ones. For a positive representative $c$, the weighted mean minimizes squared deviations identity givesThus $c=G_w$ uniquely minimizes weighted quadratic loss in logarithmic coordinates. The weighted arithmetic mean minimizes squared discrepancies in the original coordinates. These two objectives explain why different choices of what constitutes an error select different summaries.

The strictly concave function $\log x$ and the reversed Jensen's inequality give $G_w\le A_w$, where $A_w={\sum }_i{w}_i{x}_i$ is the weighted arithmetic mean. Apply the same reasoning to the multiplicative inverses to get $H_w\le G_w$ for the weighted harmonic mean. Thus $H_w\le G_w\le A_w$; equality holds exactly when all positive-weight inputs agree. For this list, $H_w=48/13$, $G_w=3\sqrt{2}$, and $A_w=21/4$. The weights specify which inputs are represented and must stay the same across the comparison.

Starting from $0<a_1\le b_1$, replace the lower endpoint by the geometric mean and the upper endpoint by the arithmetic mean. The arithmetic-geometric mean inequality gives $a_n\le a_{n+1}\le b_{n+1}\le b_n$. The bounded monotone sequence theorem gives limits, and the arithmetic recurrence forces them to coincide. The limiting value is the arithmetic-geometric mean of the initial pair.