# Taming the AGM

This post is in response to Peter’s post introducing the Approximate Geometric Mean.

The approximate geometric mean $\mathrm{(AGM)}$ is a nice approximation of the geometric mean $\mathrm{(GM)}$, but it has some quirks as we will see. After a discussion at the MathsJam gathering, I was intrigued to find out how good an approximation it is.

To get a better understanding, we first have to look again at its definition. For $A=a\cdot 10^x$ and $B=b \cdot 10^y$, we set

$\mathrm{AGM}(A,B):=\mathrm{AM}(a,b)\cdot 10^{\mathrm{AM}(x,y)}$

where $\mathrm{AM}$ stands for the arithmetic mean. This makes also sense when $a$ and $b$ are not just integers between 1 and 10, but any real numbers. Note that we won’t consider negative $A$ and $B$ (i.e. negative $a$ and $b$), as the geometric mean runs into issues if we do so. The values of $x$ and $y$ may be negative, though. The $\mathrm{AGM}$ looks like a mix between the $\mathrm{AM}$ and the $\mathrm{GM}$, so what can possibly go wrong?

### Same mean, different numbers

In contrast to the $\mathrm{AM}$ and the $\mathrm{GM}$, the $\mathrm{AGM}$ depends on the number base (10 in this case) and the presentation of $A$ and $B$.

If we write $A=(10a) \cdot 10^{(x-1)}$, we get a different value for $\mathrm{AGM}(A,B)$. This looks rather unfortunate, but it will turn out to be helpful. To ease notation we will assume in the following that $a\geq b$ unless otherwise stated. This can be done without loss of generality as $\mathrm{AGM}(A,B)=\mathrm{AGM}(B,A)$.

Peter Rowlett proved in his post that $\mathrm{GM}\leq \mathrm{AGM}$. The question is, how far can the $\mathrm{AGM}$ exceed the $\mathrm{GM}$? In other words, what’s the supremum of the ratio $R=\mathrm{AGM}/\mathrm{GM}$?

Using the notation of $A$ and $B$ as above we get
\begin{align*} R=\frac{\mathrm{AGM}(A,B)}{\mathrm{GM}(A,B)}= \frac{\mathrm{AM}(a,b)}{\mathrm{GM}(a,b)} = \frac{1}{2}\cdot (\sqrt{a/b}+\sqrt{b/a}).\end{align*}

So, the ratio $R$ doesn’t depend on $x$ and $y$ but only on $a$ and $b$. That’s convenient. Taking $a$ and $b$ in the interval $[1,10)$, as is usual, we can look at the plot of $R(a,b)$.

As long as we are in the blue part of the graph, $\mathrm{AGM}$ looks to be a sensible approximation of the $\mathrm{GM}$. So let’s look at the bad combinations of $a$ and $b$.

The worst case happens when $a$ and $b$ are maximally far apart: The supremum of $R(a,b)$ is its limit for $a \rightarrow 10$ and $b=1$. So in general, $1\leq R \lt 5.5/\sqrt{10} \approx 1.74$.

This supremum doesn’t look too bad at first, but unfortunately, the result can be unusable in extreme cases. For example, if $a=999=9.99\cdot 10^2$ and $b=1000=1 \cdot 10^3$, we have $\mathrm{GM}(A,B)\approx \mathrm{AM}(A,B)=999.5$ and $\mathrm{AGM}(A,B)\approx 1738$ – not only is the $\mathrm{AM}$ a better approximation of the $\mathrm{GM}$ than the $\mathrm{AGM}$ in this instance, the $\mathrm{AGM}$ is bigger than both the numbers $A$ and $B$ of which it is supposed to give some kind of mean!

Let’s analyse this a bit deeper. The ratio $R$ only depends on the ratio $r=a/b$. In closed form we can write $R(r)=1/2\cdot (\sqrt{r}+\sqrt{r}^{-1})$ and we are left to study this function in the range $[1,10]$. Its maximum is $R(10)$, but smaller $r$ give better results. And we will see, that we don’t have to put up with $r=10$.

Here, the flexibility of the definition of the $\mathrm{AGM}$ comes into play. Due to the choice of a suitable presentation of the numbers we can guarantee that $r$ isn’t too big. If we have $r\leq \sqrt{10}$ which is equivalent to $\sqrt{10}b\geq a \geq b$ we calculate $\mathrm{AGM}(A,B)$ as above. If $a>\sqrt{10}b$, we change the presentation of the number:

\begin{align*}B=b \cdot 10^y = (10b)\cdot 10^{y-1}=:b’ \cdot 10^{y-1} \end{align*}
and continue from there.

So, let’s redefine the $\mathrm{AGM}$ for $10>a\geq b\geq 1$ like this:

$\mathrm{AGM}(A,B)=\begin{cases} \mathrm{AM}(a,b)\cdot 10^{\mathrm{AM}(x,y)}, & \sqrt{10}b\geq a,\\ \mathrm{AM}(10b,a)\cdot 10^{\mathrm{AM}(x,y-1)}, & \text{otherwise}. \end{cases}$

Note, that in the second case we have $\sqrt{10}a>10b>a$, so that the roles of the pair $(a,b)$ are taken over by the pair $(10b,a)$. Setting $r=10b/a$ in the second case, we have in both cases $1\leq r\leq \sqrt{10}$, so we only have to study $R(r)$ in the interval $[1,\sqrt{10}]$, which will turn out to be rather benign.

Note also, that this new $\mathrm{AGM}$ can still be calculated without a calculator when using the approximation $\sqrt{10}\approx 3$, as Colin Beveridge suggested in Peter’s post.

In the example above with $A=999$ and $B=1000$ we write $B=10\cdot 10^2$ and find with this new definition of the $\mathrm{AGM}$:

\begin{align*}\mathrm{AGM}(999;1000)=\mathrm{AM}(9.99;10) \cdot 10^{\mathrm{AM}(2;2)}=999.5, \end{align*}

This coincides with the arithmetic mean of the two numbers and is really close to the geometric mean. This is looking promising.

If we define the $\mathrm{AGM}$ of two numbers $A$ and $B$ in the way explained above, we get the following two inequalities:

\begin{align*} (I) \quad & \mathrm{GM}(A,B)\leq \mathrm{AGM}(A,B) \leq \mathrm{GM}(A,B) \cdot 1.2 \\ (II) \quad & \mathrm{GM}(A,B) \leq \mathrm{AGM}(A,B) \leq \mathrm{AM}(A,B) \end{align*}Both inequalities together mean that not a lot can go wrong when using the $\mathrm{AGM}$ with the appropriate presentation of the numbers: The $\mathrm{AGM}$ is bigger than the $\mathrm{GM}$, but exceeds it by maximally 20%, and it is always smaller than the $\mathrm{AM}$.

As a consequence, the $\mathrm{AGM}$ will always be between $A$ and $B$. So it is indeed a “mean” of some kind.

### A proof of these two inequalities

(I) We only have to find the maximum of $R=\mathrm{AGM}/\mathrm{GM}$. Due to the discussion above we can assume that $\sqrt{10}b\geq a \geq b$, but $a$ can now be bigger than 10. The latter is not a problem though.

The maximum of $R=\mathrm{AGM}(A,B)/\mathrm{GM}(A,B)=\mathrm{AM}(a,b)/\mathrm{GM}(a,b)$ is attained when $a$ and $b$ are maximally apart, i.e. $r=\sqrt{10}$, so
$\max\left(\frac{\mathrm{AGM}(A,B)}{\mathrm{GM}(A,B)}\right)=\frac12 \cdot (10^{1/4}+{10^{-1/4}}) \approx 1.2.$

(II) We will show that $\mathrm{AGM}(A,B)/\mathrm{AM}(A,B) \leq 1$. Let’s drop the assumption that $a\geq b$. Instead we assume, again without loss of generality, that $x\geq y$, so that we can set $z=:x-y \geq 0$. For the ratio $r=a/b$ we have $\sqrt{10} \geq r \geq 1/\sqrt{10}$. If $r$ fell outside this interval, we would have had to change the presentation of one of the numbers before calculating the $\mathrm{AGM}$. Dividing the numerator and denominator in the above inequality by $B$ we get:
$\frac{\mathrm{AGM}(A,B)}{\mathrm{AM}(A,B)}=\frac{(1+r) \cdot 10^{z/2}} {1+r \cdot 10^z}.$

So we look for an upper bound of the function $f_z(r):=\frac{(1+r) \cdot 10^{z/2}} {1+r \cdot 10^z}$ when varying $z$ and $r$ and want to show that this upper bound is smaller or equal to 1. Note, that we only have to check for integer $z\geq 0$ (The result is actually false if we allow any real $z$).

For $z=0$, we have $f_0(r)=1$ for any $r$ and hence $\mathrm{AGM}=\mathrm{AM}$. For a fixed $z \geq 1$ we can derive the function $f_z(r)$ with respect to $r$ and find that the slope is always negative. Hence for a fixed $z$, the function $f_z(r)$ attains a maximum when $r$ is smallest, i.e. $r=1/\sqrt{10}$, so we are left to show that

$f_z(1/\sqrt{10})=\frac{(1+10^{-1/2})10^{z/2}}{1+10^{z-1/2}}\leq 1.$

For $z=1$ we have equality again and $\mathrm{AGM} = \mathrm{AM}$. For $z\geq 2$ we can write $z=2+z’$ with $z’$ being an integer $\geq 0$. We get the following chain of inequalities

$f_z\left(\frac{1}{\sqrt{10}}\right)=\frac{(1+10^{-1/2})10^{(2+z’)/2}}{1+10^{3/2 + z’}}<\frac{2\cdot 10^{1+z’/2}}{10^{3/2+z’}}\leq \frac{2}{\sqrt{10}}<1.$

This proves the second inequality. ☐

In summary, modifying the definition of the $\mathrm{AGM}$ to assure that the ratio of the “leading characters” is as close to 1 as possible, makes sure that the $\mathrm{AGM}$ works well, even in the bad cases.

# Approaching Fermi problems with the approximate geometric mean

I gave a talk on Fermi problems and a method for approaching them using the approximate geometric mean at the Maths Jam gathering in 2017. This post is a write up of that talk with some extras added in from useful discussion afterwards.

Enrico Fermi apparently had a knack for making rough estimates with very little data. Fermi problems are problems which ask for estimations for which very little data is available. Some standard Fermi problems:

• How many piano tuners are there in New York City?
• How many hairs are there on a bear?
• How many miles does a person walk in a lifetime?
• How many people in the world are talking on their mobile phones right now?

Hopefully you get the idea. These are problems for which little data is available, but for which intelligent guesses can be made. I have used problems of this type with students as an exercise in estimation and making assumptions. Inspired by a tweet from Alison Kiddle, I have set these up as a comparison of which is bigger from two unknowable things. Are there more cats in Sheffield or train carriages passing through Sheffield station every day? That sort of thing.

# Ten years and eight days

On 31st January 2008, I gave my first lecture. I was passing my PhD supervisor in the corridor and he said “there might be some teaching going if you fancy it, go and talk to Mike”. And that, as innocuous as it sounds, was the spark that lit the flame. I strongly disliked public speaking, having hardly done it (not having had much chance to practice in my education to date – I may have only given one talk in front of people to that point, as part of the assessment of my MSc dissertation), but I recognised that this was something I needed to get over. I had just started working for the IMA, where my job was to travel the country giving talks to undergraduate audiences, and I realised that signing up to a regular lecture slot would get me some much-needed experience. I enjoyed teaching so much that I have pursued it since.

I just noticed that last Wednesday was ten years since that lecture. It was basic maths for forensic science students. I was given a booklet of notes and told to either use it or write my own (I used it), had a short chat about how the module might work with another lecturer, and there I was in front of the students. That was spring in the academic year 2007/8 and this is the 21st teaching semester since then. This one is the 15th semester during which I have taught — the last 12 in a row, during which I got a full-time contract and ended ten years of part-time working.

I have this awful feeling this might lead people to imagine I’m one of the people who knows what they are doing.

P.S. The other thing that I started when I started working for the IMA was blogging – yesterday marks ten years since my first post. So this post represents the start of my second ten years of blogging.

# Carnival of Mathematics 154

The next issue of the Carnival of Mathematics, rounding up blog posts from the month of January, and compiled by Rachel, is now online at The Math Citadel.

The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

# Gerrymandering Gives Mathematics’s Moon a Day in the Sun

If you pay attention to United States politics you have probably noticed that mathematics is currently enjoying a rare moment of relevance. You probably also know this is not happening because all of a sudden politicians have decided that mathematics is clearly the coolest thing in the world, even though it clearly is, but instead because gerrymandering has become one of the major issues du jour.

# Review: Closing the Gap, by Vicky Neale

Did you read Cédric Villani’s Birth of a Theorem? Did you have the same reaction as me, that all of the mentions of the technical details were incredibly impressive, doubtless meaningful to those in the know, but ultimately unenlightening?

Writing about maths, especially deep technical maths, so that a reader can follow along with it is hard – the Venn diagram of the set of people who can write clearly and the set of people who understand the maths, two relatively small sets, has a yet smaller intersection.

# Are you more likely to be killed by a meteor or to win the lottery?

This tweet from the QI Elves popped up on my Twitter timeline:

In the account’s usual citationless factoid style, the Elves state that you’re more likely to be crushed by a meteor than to win the jackpot on the lottery.

The replies to this tweet were mainly along the lines of this one from my internet acquaintance Chris Mingay:

Yeah, why don’t we hear about people being squished by interplanetary rocks all the time? I’d tune in to that!