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 ⨉ 106
Adding 4.5 years is meaningless
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)
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 \(y \in [y_1, y_2]\).
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 \[\begin{aligned} \frac{1}{[y_1, y_2]} &= \left[\tfrac{1}{y_2}, \tfrac{1}{y_1} \right] \textrm{ if }\;0 \notin [y_1, y_2]\\ \frac{1}{[y_1, 0]} &= \left[-\infty, \tfrac{1}{y_1} \right] \end{aligned}\]
\[\begin{aligned} \frac{1}{[0, y_2]} &= \left [\tfrac{1}{y_2}, \infty \right ] \\ \frac{1}{[y_1, y_2]} &= \left [-\infty, \tfrac{1}{y_1} \right ] \cup \left [\tfrac{1}{y_2}, \infty \right ] \textrm{ if }\;0 \in (y_1, y_2) \end{aligned}\]
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 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
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 ± 4cm3
(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
\[ density=[2.4, 4.1] \frac{gr}{cm^3} \]