(or in the back of an envelope)

At first, we go by powers of ten \[10^{-1}, 10^{0}, 10^{1}, 10^{2}, 10^{3}\] This is very “low resolution”

To do better, we can increment the exponent by 0.5 \[10^{-1}, 10^{-0.5}, 10^{0}, 10^{0.5}, 10^{1}\]

Since \(10^{0.5} = \sqrt{10}≈ 3.16≈3\) we can say \[0.1, 0.3, 1, 3, 10, 30, 100,…\]

To have higher resolution, we can combine two guesses

- \(A\): A power of ten that is maybe too small
- \(B\): A power of ten that is maybe too big

Since we are using *powers*, the “average” is the
*geometric mean* \[C=\sqrt{A⋅B}\] because \[\log C=\frac{\log A + \log B}2\]

This can be approximated taking the average of the mantisas and the average of the exponents

If \(A=a× 10^x\) and \(B=b×10^y\) then \[\begin{aligned} \sqrt{A⋅B}&=\sqrt{a⋅b×10^{x+y}}\\ &=\sqrt{a⋅b}×10^{\frac{x+y}2}\\ &\approx \frac{a+b}{2}×10^{\frac{x+y}2} \end{aligned} \] when \(a\) and \(b\) are between 1 and 9

In 2016 a large stone ball was found in Podubravlje village near Zavidovici, Bosnia and Herzegovina.

We do not know how these stones were made

Similar stones have been found in Costa Rica

Nevertheless, people have made small round stones to sell as souvenirs

I was given a stone ball from Bosnia

We want to know its density

So we need to know mass and volume

- Estimate its mass
- Estimate its volume
- Estimate its density

**Ask questions**

Based on intuition we think that the mass is more than 100gr and less than 1Kg

Taking the geometric mean, we got 300gr

But it can be anything between 200gr and 400gr

We write 300gr ± 100gr

Comparing with a tea cup, we guess 80ml

But it can be anything between 70ml and 90ml

We write 80ml ± 10ml

**How would you estimate the density?**

We will represent measurements as intervals \([x_1, x_2]\)

A binary operation \(\star\) on two intervals is defined by

\[[x_1, x_2] {\,\star\,} [y_1, y_2] = \{ x \star y \, | \, x \in [x_1, x_2] \text{ and } y \in [y_1, y_2] \}.\]

In other words, it is the set of all possible values of \(x \star y\),

where \(x\) and \(y\) are in their corresponding
intervals.

Wikipedia: “Interval arithmetic”

If \(\star\) is either \(+, -, \cdot,\) or \(÷\), then \([x_1, x_2] \star [y_1, y_2]\) is \[ [\min\{ x_1 \star y_1, x_1 \star y_2, x_2 \star y_1, x_2 \star y_2\},\\ \max \{x_1 \star y_1, x_1 \star y_2, x_2 \star y_1, x_2 \star y_2\} ], \] as long as \(x \star y\) is defined for all \(x\in [x_1, x_2]\) and \(y \in [y_1, y_2]\).

- Addition: \[[x_1, x_2] + [y_1, y_2] = [x_1+y_1, x_2+y_2]\]
- Subtraction: \[[x_1, x_2] - [y_1, y_2] = [x_1-y_2, x_2-y_1]\]

This is the area of a rectangle with varying edges

\([x_1, x_2] \cdot [y_1, y_2]\) is \[[\min \{x_1 y_1,x_1 y_2,x_2 y_1,x_2 y_2\},\\ \max\{x_1 y_1,x_1 y_2,x_2 y_1,x_2 y_2\}]\]

The result interval covers all possible areas, from smallest to the largest

\[\frac{[x_1, x_2]}{[y_1, y_2]} = [x_1, x_2] \cdot \frac{1}{[y_1, y_2]},\] where \[\begin{aligned} \frac{1}{[y_1, y_2]} &= \left[\tfrac{1}{y_2}, \tfrac{1}{y_1} \right] \textrm{ if }\;0 \notin [y_1, y_2]\\ \frac{1}{[y_1, 0]} &= \left[-\infty, \tfrac{1}{y_1} \right] \end{aligned}\]

\[\begin{aligned} \frac{1}{[0, y_2]} &= \left [\tfrac{1}{y_2}, \infty \right ] \\ \frac{1}{[y_1, y_2]} &= \left [-\infty, \tfrac{1}{y_1} \right ] \cup \left [\tfrac{1}{y_2}, \infty \right ] \textrm{ if }\;0 \in (y_1, y_2) \end{aligned}\]

- Exponential function: \(a^{[x_1, x_2]} = [a^{x_1},a^{x_2}]\) for \(a > 1,\)
- Logarithm: \(\log_a [x_1, x_2] = [\log_a {x_1}, \log_a {x_2}]\) for positive intervals \([x_1, x_2]\) and \(a>1,\)
- Odd powers: \([x_1, x_2]^n = [x_1^n,x_2^n]\), for odd \(n\in ℕ\)

What happens in *even* powers?

Let’s measure the diameter using a caliper

It is about 5.3cm

Since the ball is not 100% spherical, the diameter varies between 5.2cm and 5.4cm

Now we can calculate the volume using the formula \[\frac{4}{3} π r^3\]

Using the central value

`[1] 2.65`

`[1] 77.95181`

Maybe the diameter is 5.2cm. Then

`[1] 73.62218`

Maybe the diameter is 5.4cm. Then

`[1] 82.44796`

The real volume is somewhere in the range

`[1] 73.62218 82.44796`

That is, an interval with center

`[1] 78.03507`

and width

`[1] 4.41289`

Notice that

`[1] 78.03507`

is not the same as

`[1] 77.95181`

but both values are close *(why?)*

For now we take the first one

We can write 78.0350671 ± 4.4128905 cm^{3}

But then, most of the decimals are meaningless

We get a false feeling of precision,

but we really do not know all the decimals

First, we round the *error term* to a single digit

`[1] 4`

Then we discard all decimals smaller than the error term

`[1] 78`

so we write 78 ± 4cm^{3}

(we get the same result if we use `vol`

)

We have \[\text{vol}=78\text{cm}^3 ± 4\text{cm}^3\] and \[\text{mass}=250\text{gr} ± 50\text{gr}\]

Now we can calculate the density

We will consider all possible cases

\[\begin{aligned} \text{vol} &=[74, 82]\text{cm}^3\\ \text{mass}&=[200, 300]\text{gr}\\ \text{density}_{\min} &=\min \left\{\frac{200}{74}, \frac{300}{74},\frac{200}{82},\frac{300}{82}\right\}\frac{\text{gr}}{\text{cm}^3}\\ \text{density}_{\max} &=\max\left\{\frac{200}{74}, \frac{300}{74},\frac{200}{82},\frac{300}{82}\right\}\frac{\text{gr}}{\text{cm}^3}\ \end{aligned} \]

\[\text{density}=[2.4390244, 4.0540541] \frac{\text{gr}}{\text{cm}^3}\] Rounding, we get \[\text{density}=[2.4, 4.1] \frac{\text{gr}}{\text{cm}^3}\]

- Train your instinct
- If you can, think slow
- Every measurement is an interval
- All calculations yield an interval
- Find the margin of error
- Omit decimals smaller than the margin of error
- Intervals can be written \([x_{min}, x_{max}]\) or \(x_{mean}±\Delta x\)

\[N = R_* \cdot f_\mathrm{p} \cdot n_\mathrm{e} \cdot f_\mathrm{l} \cdot f_\mathrm{i} \cdot f_\mathrm{c} \cdot L\]

- \(R_{*}\) = 1 yr
^{-1} - \(f_{p}\) = 0.2 to 0.5
- \(n_{e}\) = 1 to 5
- \(f_{l}\) = 1
- \(f_{i}\) = 1
- \(f_{c}\) = 0.1 to 0.2
- \(L\) = 1000 to 100,000,000