# Other iterated means - AHM, GHM.
[[am-gm-agm|Previously]] we discussed the arithmetic-geometric mean of two positive reals, $\operatorname{agm}(x,y)$, which can be found by constructing two sequences of the arithmetic mean and the geometric mean. Namely $$
x_{0} = x, y_{0} = y, x_{n+1} = a(x_{n}, y_{n}), y_{n+1} = g(x_{n},y_{n})
$$where $a(x,y) = \frac{1}{2}(x+y)$, $g(x,y) = \sqrt{xy}$, then $\operatorname{agm}(x,y) = \lim x_{n} = \lim y_{n}$.
What if instead we used different means instead to construct these sequences? Recall the **harmonic mean** of two positive reals is given by $$
h(x,y) = \frac{2}{\frac{1}{x}+\frac{1}{y}},
$$the reciprocal of the average of the reciprocals of $x$ and $y$.
We have the following classic result,
> For $x, y > 0$, we have $$
\min(x,y) \le h(x,y) \le g(x,y) \le a(x,y) \le \max(x,y)
$$
$\blacktriangleright$ Indeed, note $x,y \ge \min(x,y)$, so $$
\frac{1}{x}+\frac{1}{y} \le \frac{2}{\min(x,y)} \implies \min(x,y) \le h(x,y).
$$
And note by invoking AM-GM itself, we have $$
\frac{1}{2} \left( \frac{1}{x}+\frac{1}{y} \right) \ge \sqrt{\frac{1}{x} \frac{1}{y}},
$$whence $$
\frac{2}{\frac{1}{x}+\frac{1}{y}} \le \sqrt{xy}
$$as desired. $\blacksquare$
With this in mind, let us define the arithmetic-harmonic mean ($\operatorname{ahm}$) and the geometric-harmonic mean ($\operatorname{ghm}$) in a similar fashion to the arithmetic-geometric mean.
## Arithmetic-harmonic mean.
For positive $x,y > 0$, define two sequences where $x_{0} = x$, $y_{0} = y$ and $$
x_{n+1} = a(x_{n},y_{n}),\quad y_{n+1} = h(x_{n},y_{n}).
$$
With the same argument as why the $\operatorname{agm}(x,y)$ exist, we see that the limits $$
\lim x_{n} = \lim y_{n} = \eta
$$exist and agree with other, since $x_{n}$ will be a monotonic decreasing sequence bounded below by $\min(x,y)$, while $y_{n}$ is a monotonically increasing sequence bounded above by $\max (x,y)$, and that $|x_{n}-y_{n}| \le \frac{1}{2^{n}}|x-y|$. Define this common limit $\eta$ as the **arithmetic-harmonic mean** of $x,y$, and denote it as $\operatorname{ahm}(x,y) =\eta$.
Again we have similar properties, for positive $x,y > 0$
- $h(x,y) \le \operatorname{ahm}(x,y) \le a(x,y)$
- $\operatorname{ahm}(x,y) = \operatorname{ahm}(y,x)$
- $\operatorname{ahm}(x,y) = \operatorname{ahm}(a(x,y), h(x,y))$
- $\operatorname{ahm}(\lambda x, \lambda y) = \lambda \operatorname{ahm}(x,y)$ for $\lambda > 0$
- $\operatorname{ahm}(x,x)=x$.
So, is there a closed form for the $\operatorname{ahm}(x,y)$? Yes, and perhaps surprisingly,
> For any positive $x,y > 0$, $$
\operatorname{ahm}(x,y) = g(x,y).
$$
$\blacktriangleright$ To see this, first observe that our sequence $$
y_{n+1} = h(x_{n},y_{n}) = \frac{2}{\frac{1}{x_{n}} + \frac{1}{y_{n}}}= \frac{2x_{n}y_{n}}{x_{n}+y_{n}} = \frac{x_{n}y_{n}}{a(x_{n},y_{n})} = \frac{x_{n}y_{n}}{x_{n+1}},
$$hence $$
x_{n+1}y_{n+1} = x_{n}y_{n} = xy.
$$And note we have $$
x_{n+1} = a(x_{n},y_{n}) = \frac{1}{2} (x_{n}+y_{n}) = \frac{1}{2}\left( x_{n}+\frac{xy}{x_{n}} \right)
$$
So as $\lim x_{n} = \eta = \operatorname{ahm}(x,y)$, as $n\to \infty$ we have equality $$
\eta = \frac{1}{2}\left( \eta + \frac{xy}{\eta} \right) \implies \eta = \sqrt{xy} = g(x,y)
$$as claimed! $\blacksquare$
So the arithmetic-harmonic mean is the geometric mean!
## Geometric-harmonic mean.
What about the geometric-harmonic mean? That is, if we define for $x ,y > 0$, the sequence $x_{0} = x$ and $y_{0} = y$ with $$
x_{n+1} = g(x_{n},y_{n}),\quad y_{n+1} = h(x_{n},y_{n}),
$$by the same arguments as before, we see that $x_{n}$ and $y_{n}$ are both monotonic, and that $|x_{n}-y_{n}|\le \frac{1}{2^{n}}|x-y|$. So these sequences converge to a common limit, say $\lim x_{n} = \lim y_{n} = \gamma$, which we can define it as the **geometric-harmonic mean** of $x$ and $y$, denote as $\operatorname{ghm}(x,y)$.
We again have similar properties, for $x,y > 0$,
- $h(x,y) \le \operatorname{ghm}(x,y) \le g(x,y)$
- $\operatorname{ghm}(x,y) = \operatorname{ghm}(y,x)$
- $\operatorname{ghm}(\lambda x, \lambda y) = \lambda \operatorname{ghm}(x,y)$ for $\lambda > 0$
- $\operatorname{ghm}(x,x) = x$.
So, we ask again, is there a closed form for $\operatorname{ghm}(x,y)$, or is it related to other means? Yes, as it turns out
> For any $x , y > 0$, we have $$
\operatorname{ghm}(x,y) = \frac{xy}{\operatorname{agm}(x,y)}.
$$
$\blacktriangleright$ Defining $x_{n+1} = g(x_{n},y_{n})$, $y_{n+1} = h(x_{n},y_{n})$, $x_{0} = x$ and $y_{0} = y$, let us define two new sequences $$
a_{n} = \frac{1}{x_{n}}, \quad b_{n} = \frac{1}{y_{n}}
$$as well, where $a_{0} = \frac{1}{x}$ and $b_{0} = \frac{1}{y}$.
Note that the sequence $a_{n}$ and $b_n$ satisfy $$
a(a_{n},b_{n}) = \frac{1}{2}(a_{n}+b_{n}) = \frac{1}{2}\left( \frac{1}{x_{n}} + \frac{1}{y_{n}} \right) = \frac{1}{h(x_{n},y_{n})} = \frac{1}{y_{n+1}} = b_{n+1}
$$and that $$
g(a_{n},b_{n}) = \sqrt{a_{n}b_{n}}=\frac{1}{\sqrt{x_{n}y_{n}}} = \frac{1}{x_{n+1}} = a_{n+1}
$$So we have $$
\operatorname{agm}\left( \frac{1}{x}, \frac{1}{y} \right) = \operatorname{agm}(a_{n},b_{n}) = \lim a_{n} = \lim \frac{1}{x_{n}} = \frac{1}{\operatorname{ghm}(x,y) },
$$so by scaling both sides by $xy$, we get $$
xy \operatorname{agm}\left( \frac{1}{x}, \frac{1}{y} \right) = \frac{xy}{\operatorname{ghm}(x,y)} \implies \operatorname{ghm}(x,y) = \frac{xy}{\operatorname{agm}(x,y) }
$$as claimed! $\blacksquare$
As a corollary, we have integral form
> For $x,y > 0$, $$
\operatorname{ghm}(x,y) = \frac{2xy}{\pi} \int_{0}^{\pi / 2} \frac{dt}{\sqrt{x^{2}\cos^{2}(t)+y^{2}\sin^{2}(t)}}.
$$
Neat!