Estimating order of magnitude

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 \[0.1, 0.3, 1, 3, 10, 30, 100,…\]

Fine tuning estimations

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 \[\log C=\frac{\log A + \log B}2\]

Easy approximation of geometric mean

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

Density of the stone ball

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

Our estimation of mass

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

Our estimation of volume

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?

Definitions

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.

Simplification for basic operations

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 \(y \in [y_1, y_2]\).

Even easier

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

Multiplication

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

Division needs more attention

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

finally

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

Functions

• 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?

Measuring the diameter

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 \[\frac{4}{3} π r^3\]

Calculating the volume

Using the central value

diam <- 5.3
r <- diam/2
r

[1] 2.65

vol <- 4/3 * pi * r^3
vol

[1] 77.95181

Being pessimistic

Maybe the diameter is 5.2cm. Then

diam_min <- 5.2
r_min <- diam_min/2
vol_min <- 4/3 * pi * r_min^3
vol_min

[1] 73.62218

Being optimistic

Maybe the diameter is 5.4cm. Then

diam_max <- 5.4
r_max <- diam_max/2
vol_max <- 4/3 * pi * r_max^3
vol_max

[1] 82.44796

A range of possibilities

The real volume is somewhere in the range

c(vol_min, vol_max)

[1] 73.62218 82.44796

That is, an interval with center

(vol_max+vol_min)/2

[1] 78.03507

and width

(vol_max-vol_min)/2

[1] 4.41289

Two possible centers

Notice that

(vol_max+vol_min)/2

[1] 78.03507

is not the same as

vol

[1] 77.95181

but both values are close (why?)

For now we take the first one

How many decimals?

We can write 78.0350671 ± 4.4128905 cm³

But then, most of the decimals are meaningless

We get a false feeling of precision,
but we really do not know all the decimals

Rounding numbers

First, we round the error term to a single digit

round((vol_max-vol_min)/2)

[1] 4

Then we discard all decimals smaller than the error term

round((vol_max+vol_min)/2)

[1] 78

so we write 78 ± 4cm³

(we get the same result if we use vol)

Each measurement is an interval

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

Intervals

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

Calculating

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

Summary

• 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 \(x_{mean}±\Delta x\)

Apply to Drake equation

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

• \(R_{*}\) = 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