Some tourists in the Museum of Natural History are marveling at some dinosaur bones. One of them asks the guard, “Can you tell me how old the dinosaur bones are?”

The guard replies, “They are 3 million, four years, and six months old.”

“That’s an awfully exact number,” says the tourist. “How do you know their age so precisely?”

The guard answers, “Well, the dinosaur bones were three million years old when I started working here, and that was four and a half years ago.”

Lets be honest about what we know and what we do not know

We write the values that have real meaning

3 million years means 3±0.5 ⨉ 10^{6}

Adding 4.5 years is meaningless

- All non-zero digits are significant
- In 1234 all digits are significant
- Same in 12.34 and 1.234

- Zeros surrounded by non-zero are significant
- Same in 1204. Four significant digits

- Zeros to the left are not significant
- Four significant digits in 0.0001234

- Zeros to the right are not significant
**unless**there is a decimal point- 12340 has four significant digits
- 123.40 has five significant digits
- 1234.0 has five significant digits
- 12340. has five significant digits

Notice that 1234.0 is not the same as 1234

Also, 12340. is not the same as 12340

This is a *convention* but not everybody uses it

It is much safer to use scientific notation

It is safer to represent numbers in *scientific notation*
\[
\begin{aligned}
1234.0 & = 1.2340\cdot 10^3\\
1234 & = 1.234 \cdot 10^3\\
12340. & = 1.2340 \cdot 10^4\\
12340 & = 1.234 \cdot 10^4
\end{aligned}
\]

All digits in the *mantissa* are significant

(*mantissa* is the number being multiplied by 10 to the power
of the *exponent*)

- Last class we measured the volume of the stone ball
- Round the uncertainty to a single figure
- Instead of \(2013.765 ± 78.345\) write \(2013.765 ± 80\)

- Then round the main value to the last well known place
- Instead of \(2013.765\) ± 80 write \(2010 ± 80\)

- The digits “3.765” were a
*white lie*. Let’s not fool ourselves

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 \[vol=78cm^3 ± 4cm^3\] and \[mass=250gr ± 50gr\]

Now we can calculate the density

We will consider all possible cases

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

\[ density=[2.4390244, 4.0540541] \frac{gr}{cm^3} \]

Rounding, we get

\[ density=[2.4, 4.1] \frac{gr}{cm^3} \]