# AM, GM, and the AGM.
In this article we discuss the **arithmetic-geometric mean** of two positive numbers $x,y > 0$, denote as $\operatorname{agm} (x,y)$, and show that we have an integral form $$
\operatorname{agm}(x,y) = \frac{\pi}{2} \left( \int_{0}^{\pi / 2} \frac{dt}{\sqrt{x^{2}\cos^{2} + y^{2}\sin^{2}(t)}} \right)^{-1}
$$
## Arithmetic mean, geometric mean, and the arithmetic-geometric mean.
Given any two non-negative reals $x,y$, define their *arithmetic mean* to be $$
a(x,y) = \frac{1}{2} (x+y)
$$and their *geometric mean* to be $$
g(x,y) = \sqrt{xy}.
$$
It is well-known that we have
> **AM-GM inequality.**
>For any two non-negative reals $x,y \ge 0$, we have inequality $$
\min(x,y) \le g(x,y) \le a(x,y) \le \max(x,y).
$$
$\blacktriangleright$ Indeed, $\min(x,y)^{2} \le xy$ and $x + y \le 2 \max(x,y)$ for $x,y \ge 0$. And note that $$
\begin{align*}
(x-y)^{2} & \ge 0 \\
\implies x^{2} - 2xy + y^{2} & \ge 0 \\
\implies x^{2} + 2xy + y^{2} & \ge 4xy \\
\implies (x+y)^{2} & \ge 4xy \\
\implies a(x,y) & \ge g(x,y). \quad\blacksquare
\end{align*}
$$
Now, take any $x,y > 0$, define a pair of sequences $(x_{n})$ and $(y_{n})$ where $x_{0} = x$, $y_{0}= y$ and $$
\begin{align*}
x_{n+1} &= a(x_{n},y_{n}) = \frac{1}{2}(x_{n}+y_{n}) \\
y_{n+1} &= g(x_{n},y_{n}) = \sqrt{x_{n}y_{n}}
\end{align*}
$$for all $n\ge 0$. That is, we repeatedly take the arithmetic mean and the geometric mean of the two previous terms.
We claim the following:
> For any $x,y > 0$, both sequences $(x_{n})$ and $(y_{n})$ converges, and in fact they converge to the same limit $\mu = \lim x_{n} = \lim y_{n}$.
$\blacktriangleright$ Let $(x_{n})$ and $(y_{n})$ be sequences defined as above. First we show both sequences converges, then we show their limits are the same.
Note after one iterate, $x_{1} = a(x,y)$ and $y_{1} = g(x,y)$, we have $y_{1} \le x_{1}$ by AM-GM inequality. And again by AM-GM inequality, we have $$
\min(x,y) \le y_{n} \le y_{n+1} \le x_{n+1} \le x_{n} \le \max(x,y)
$$for all $n\ge 1$. So $(x_{n})$ is a decreasing sequence bounded below by $y_{1}$, while $(y_{n})$ is an increasing sequence bounded above by $x_{1}$. Hence by monotone sequence theorem, the limits $\lim x_{n}$ and $\lim y_{n}$ both exist.
Now we show $\lim x_{n} = \lim y_{n}$. Let us denote $\delta_{n} = |x_{n}-y_{n}|$. Note that $x_{2},y_{2} \in [y_{1},x_{1}]$ and that $x_{2}$ is the midpoint of $x_{1}$ and $y_{1}$. Hence $\delta_{2}:= |y_{2} - x_{2}| \le \frac{\delta_{1}}{2}$. And in general $$
\delta_{n} = |x_{n} - y_{n}| \le \frac{1}{2^{n}} |x_{0}-y_{0}|.$$So $$
\lim |x_{n} - y_{n} | = 0,
$$ whence $\lim x_{n} = \lim y_{n} = \mu$, some $\mu$. $\blacksquare$
This gives a well-defined notion of the *arithmetic-geometric mean* of two non-negative reals:
> **The arithmetic-geometric mean of two non-negative reals.** For any $x,y \ge 0$, define their arithmetic-geometric mean as $\operatorname{agm}(x,y) = \mu$, where $\mu$ is the common limit of the repeated arithmetic and geometric mean sequences $(x_{n})$ and $(y_{n})$ as defined above.
Note if one of $x$ or $y$ equal zero, then $\operatorname{agm}(x,y) = 0$.
By the proofs in this construction, we see that for any two non-negative reals where $x,y \ge 0$, we have
> $$
\min(x,y) \le g(x,y) \le \operatorname{agm} (x,y) \le a(x,y) \le \max(x,y).
$$
For example, consider $x = x_{0} = 1$ and $y =y_{0}= 2$, if we compute their iterated arithmetic means and geometric means, we get a table, for all $n\ge 1$, $$
\begin{array}{c|c|c}
n & x_{n} = a(x_{n-1},y_{n-1}) & y_{n} = g(x_{n-1},y_{n-1})\\ \hline
0& 1 & 2 \\
1& 1.5 & \sqrt{2} =1.41421356237 \\
2& 1.4571067811865475 & 1.4564753151219703 \\
3 & 1.4567910481542587 & 1.456791013939555\\
4& 1.4567910310469068 & 1.4567910310469068 \end{array}
$$so $\operatorname{agm}(1,2) \approx 1.4567910310469068...$. Note this converges quite rapidly, as the difference $|x_{n}-y_{n}|$ decreases by a factor of 2 in each iteration.
Some basic properties of $\operatorname{agm}(x,y)$.
For any $x, y \ge 0$, we have
- $\operatorname{agm}(x,y) = \operatorname{agm}(y,x)$
- $\operatorname{agm}(x,y) = \operatorname{agm}(a(x,y),g(x,y))$
- $\operatorname{agm}(x,0) = 0$
- $\operatorname{agm}(\lambda x,\lambda y)=\lambda \operatorname{agm}(x,y)$, for $\lambda \ge 0$.
- $\operatorname{agm}(u,u) = u$, for all $u \ge 0$.
Note, by the homogeneity of $\operatorname{agm}(x,y)$, we can consider the function $$
f(x) = \operatorname{agm}(x,1).
$$
What is the asymptotic of $f(x)=\operatorname{agm}(x,1)$? As it turns out, we have $$
\operatorname{agm}(x,1) \sim \frac{2}{\pi} \frac{x}{\log(x)}, \text{ as } x\to\infty.
$$
(Connection to prime number theorem?? Note if we denote $\pi(x)$ as the number of primes less than or equal to $x$, then $\pi(x) \sim x / \log x$ as well.)
## Integral form of the AGM.
Is it possible to find a closed form for $\operatorname{agm}(x,y)$? In fact Gauss found an integral form!
> **Gauss integral form for the AGM.**
> For any two positive reals $x,y > 0$ we have $$
\operatorname{agm} (x,y) = \frac{\pi / 2 }{\displaystyle \int_{0}^{\pi / 2} \frac{dt}{\sqrt{x^{2} \cos^{2}(t) + y^{2}\sin^{2}(t)}}}
$$
Astounding!
In fact, at age 22 Gauss observed purely computationally, verified to 11 decimal places by hand, that $$
\frac{1}{\operatorname{agm} (1,\sqrt{2})} = \frac{2}{\pi} \int_{0}^{1} \frac{dt}{\sqrt{1-t^{4}}},
$$which he wrote in his diary, that if this fact is true, then it shall lead to new development in analysis. This value $\operatorname{agm}(1,\sqrt{2}) \approx 1.1981402347355923$, also known as *Gauss constant*, converges quite quickly.
Denote this integral $$
J(x,y) = \int_{0}^{\pi/2} \frac{dt}{\sqrt{x^{2}\cos^{2}(t)+y^{2}\sin^{2}(t)}}.
$$The strategy is to show that $$
J\left( \frac{x+y}{2}, \sqrt{xy} \right) = J(x,y).
$$
How Gauss establish the integral identity is to take a suitable substitution, but it isn't immediate to see how to come about this choice of substitution.
We will instead re-express the integral to be something more symmetric, by applying Lagrange's identity to better determine what to substitute.
Ok. Let us lay out our overall task.
Denote for $x,y > 0$ the integral $$
J(x,y) = \int_{0}^{\pi/2} \frac{dt}{\sqrt{x^{2}\cos^{2}(t)+y^{2}\sin^{2}(t)}},
$$and we wish to show that $$
\operatorname{agm} (x,y) J(x,y) = \frac{\pi}{2}.
$$
First we re-express $J(x,y)$ into something more symmetric. By factoring, we see that $$
J(x,y) = \int_{0}^{\pi/2} \frac{dt}{x\cos(t) \sqrt{1+\left( \frac{y}{x} \right)^{2}\tan^{2}(t)}}
$$which suggests the substitution $$
u = \frac{y}{x} \tan(t)
$$so $$
\arctan(x u / y) = t,\quad dt= \frac{(x / y) du}{1+(x u /y)^{2}}
$$Note as $t$ ranges from $0$ to $\pi / 2$, we have $u$ ranging from $0$ to $+\infty$. Also for $t\in [0, \frac{\pi}{2}]$, we have $\cos(t),\sin(t),\tan(t) \ge 0$, so $$
1 + \tan^{2}(t) = \frac{1}{\cos^{2}(t)}
$$and $$
\cos(t) = \frac{1}{\sqrt{1+\tan^{2}(t)}} = \frac{1}{\sqrt{1+(x u / y)^{2}}}.
$$
With these substitutions, we have now $$
\begin{align*}
J(x,y) & = \int_{u=0}^{\infty} \frac{(x / y) du}{(1 + (x u / y)^{2})} \frac{\sqrt{1 + (x u / y)^{2}}}{x \sqrt{1 + u^{2}}} \\
& = \int_{u=0}^{\infty} \frac{(1 / y) du}{\sqrt{1+ (x u / y)^{2}}\sqrt{1 + u^{2}}} \\
& = \int_{u = 0}^{\infty} \frac{du}{\sqrt{y^{2} + x^{2}u^{2}}\sqrt{1+u^{2}}}
\end{align*}
$$
If we employ another simple substitution $w = xu$, $dw = x du$, then $$
\begin{align*}
J(x,y) & = \int_{w=0}^{\infty} \frac{dw}{x \sqrt{y^{2} +w^{2}} \sqrt{1+(w / x)^{2}}} \\
&= \int_{w=0}^{\infty} \frac{dw}{\sqrt{y^{2} +w^{2}} \sqrt{x^{2}+w^{2}}}
\end{align*}
$$
So after relabeling, we establish the identity
> For $x, y > 0$, we have $$
J(x,y) = \int_0^{\pi / 2} \frac{dt}{\sqrt{x^{2}\cos^{2}(t)+y^{2}\sin^{2}(t)}} = \int_{0}^{\infty} \frac{dt}{\sqrt{(x^{2}+t^{2})(y^{2}+t^{2})}}
$$
In other words, $J(x,y)$ is an **elliptic integral**. By modern definition, an elliptic integral is an integral of the form $$
\int R\left (t,\sqrt{P(t)} \right) dt
$$where $R(t,u)$ is a rational function in $t,u$, and $P$ is a polynomial of degree 3 or 4, typically without repeated roots.
Next, using this equivalent elliptic integral form of $J(x,y)$, we will show $$
J\left( \frac{x+y}{2} , \sqrt{xy}\right) = J(x,y).
$$Note we have $$
J\left( \frac{x+y}{2}, \sqrt{xy} \right) = \int_{0}^{\infty} \frac{dt}{\sqrt{\left(\frac{x+y}{2}\right)^{2}+t^{2}}\sqrt{ xy+t^{2}}}
$$
To do this we recall **Lagrange's identity**, where $$
(A^{2} +B^{2})(C^{2}+D^{2}) = (AC + BD)^{2} + (AD - BC)^{2}.
$$There are two ways to apply Lagrange's identity to the radicand in $J(x,y)$, one way leads to $$
(x^{2}+t^{2})(y^{2}+t^{2}) = (xt + yt)^{2}+(xy-t^{2})^{2} = (x+y)^{2}t^{2} + (xy-t^{2})^{2}
$$which looks promising as we have terms like $(x+y)$ and $xy$.
With this, we have $$
\begin{align*}
J(x,y) & = \int_{0}^{\infty} \frac{dt}{\sqrt{(x+y)^{2}t^{2}+(xy-t^{2})^{2}}} \\
& = \int_{0}^{\infty} \frac{dt}{t\sqrt{(x+y)^{2}+\left( \frac{xy}{t} - t \right)^{2}}}
\end{align*}
$$
This prompts the substitution $$
u = t - \frac{xy}{t} , \quad du = \left( 1 + \frac{xy}{t^{2}} \right) dt = \left( t + \frac{xy}{t} \right) \frac{dt}{t}
$$And note when $t$ ranges from $0$ to $\infty$, we have $u$ range from $-\infty$ to $+\infty$. So $$
J(x,y) = \int_{u = -\infty}^{+\infty} \frac{du}{\left( t+\frac{xy}{t} \right)\sqrt{(x+y)^{2}+u^{2}}}
$$To express $t + \frac{xy}{t}$ in terms of $u$, we solve for $t$ in terms of $u$. Since $u = t - \frac{xy}{t}$, we have $$
tu = t^{2}-xy \implies t = \frac{u \pm \sqrt{u^{2}+4xy}}{2}
$$But since $t \ge 0$, we must have the $+$ sign in expression above. So $$
\begin{align*}
t + \frac{xy}{t}&= \frac{u}{2}+\frac{\sqrt{u^{2}+4xy}}{2} + \frac{2xy}{u+\sqrt{u^{2}+4xy}} \\
&= \frac{u}{2} + \frac{\sqrt{u^{2}+4xy}}{2} + \frac{2xy(u-\sqrt{u^{2}+4xy})}{u^{2}-(u^{2}+4xy)} \\
&= \frac{u}{2} + \frac{\sqrt{u^{2}+4xy}}{2} - \frac{u}{2} + \frac{\sqrt{u^{2}+4xy}}{2} \\
\implies t + \frac{xy}{t}&= \sqrt{u^{2}+4xy}
\end{align*}
$$Hence $$
\begin{align*}
J(x,y) & = \int_{-\infty}^{\infty} \frac{du}{\sqrt{u^{2}+4xy}\sqrt{(x+y)^{2}+u^{2}}} \\
& = 2 \int_{0}^{\infty} \frac{du}{\sqrt{u^{2}+4xy}\sqrt{(x+y)^{2}+u^{2}}}
\end{align*}
$$as the integrand is an even function in $u$. Finally, by setting $2 w = u$, $2dw = du$, we have $$
\begin{align*}
J(x,y) &= 2 \int_{0}^{\infty} \frac{2dw}{\sqrt{4w^{2}+4xy}\sqrt{(x+y)^{2}+4w^{2}}} \\
& = \int_{0}^{\infty} \frac{dw}{\sqrt{w^{2}+xy} \sqrt{\left( \frac{x+y}{2} \right)^{2}+w^{2}}} \\
\implies J(x,y) &= J\left( \frac{x+y}{2} , \sqrt{xy}\right)
\end{align*}
$$as claimed!
Finally, we establish the claim of $J(x,y)$ with $\operatorname{agm}(x,y)$. Take any $x,y> 0$, and set $x_{0} = x$, $y_{0} = y$, and $$
x_{n+1} = \frac{x_{n}+y_{n}}{2}, y_{n+1} = \sqrt{x_{n}y_{n}}
$$we have $$
J(x,y) = J(x_{n},y_{n})
$$for all $n\ge 0$.
Now, convince ourselves that $J(x,y)$ is continuous for $x,y > 0$, so as $\lim x_{n} = \lim y_{n} = \mu = \operatorname{agm}M(x,y)$, we have $$
J(x,y) = J(x_{n}, y_{n}) = \lim J(x_{n},y_{n}) = J(\mu,\mu)
$$where $\mu = \operatorname{agm}(x,y)$. And since $$
J(\mu,\mu) = \int_0^{\pi / 2} \frac{dt}{\sqrt{\mu^{2}\cos^{2}(t)+\mu^{2}\sin^{2}(t)}} = \frac{\pi}{2\mu},
$$we see that $$
J(x,y) = \frac{\pi}{2 \operatorname{agm} (x,y)}
$$as claimed! $\blacksquare$