Aristotle (384–322 BC), in “Nicomachean Ethics”

The interval \([x_{\min}, x_{\max}]\) can also be written as \[x ± Δx\] where \[x = \frac{x_{\max} + x_{\min}}{2}\qquad Δx = \frac{x_{\max} - x_{\min}}{2}\]

If the interval does not contain 0, then \(Δx\) is smaller than \(x\) (that is, if \(x_{\min}\) and \(x_{\max}\) have the same sign)

In other words, we assume \[\frac{Δx}{x} << 1\]

Sum of two measurements:

\[(x ± Δx) + (y ± Δy) = (x+y) ±
(Δx+Δy)\]

Difference between measurements:

\[(x ± Δx) - (y ± Δy) = (x-y) ±
(Δx+Δy)\]

To calculate \((x ± Δx) × (y ± Δy)\) we first write \[(x ± Δx) = x(1 ± Δx/x)\\ (y ± Δy) = y(1 ± Δy/y)\]

Then we sum the relative errors \[xy (1 ± (Δx/x + Δy/y))\]

Finally we expand the result \[xy ± xy(Δx/x + Δy/y)\]

\[ \begin{aligned} (x ± Δx) (y ± Δy) & = x(1 ± Δx/x) y(1 ± Δy/y)\\ & = xy(1 ± Δx/x)(1 ± Δy/y) \\ & = xy(1 ± Δx/x ± Δy/y ± (Δx/x)(Δy/y)) \\ & ≈ xy(1 ± Δx/x + Δy/y) \\ & = xy ± xy(Δx/x + Δy/y)\\ \end{aligned} \]

We discard \((Δx/x)(Δy/y)\) because it is small

Using the previous formula, we have \[(x ± Δx)^2 = x^2(1 ± Δx/x + Δx/x) = x^2(1 ± 2⋅Δx/x)\] Indeed \[ \begin{aligned} (x ± Δx)^2 & = x^2(1 ± Δx/x)^2\\ & = x^2(1 ± 2Δx/x + (Δx/x)^2)\\ & ≈ x^2(1 ± 2Δx/x) \\ \end{aligned} \] because \((Δx/x)^2\) is very small and can be ignored

Using the binomial theorem, we have \[(1 + Δx/x)^n = \sum_{k=0}^n {n\choose k} (Δx/x)^k\] Since \((Δx/x)^k\) is very small for all \(k>1\), we can ignore them \[ (1 + Δx/x)^n ≈ {n\choose 0} (Δx/x)^0 + {n\choose 1} (Δx/x)^1 = 1 + n(Δx/x) \]

The result is valid also for \(-Δx/x\), so we have \[\left(1 ± \frac{Δx}{x}\right)^n ≈ 1 ± n\frac{Δx}{x}\] therefore we can write \[(x ± Δx)^n =\left(x\left(1 ± \frac{Δx}{x}\right)\right)^n ≈ x^n\left(1 ± n\frac{Δx}{x}\right)\]

To calculate \(\sqrt{x+Δx}\) we need to find \(\sqrt{1+Δx/x}\)

We will assume, as we have seen before, that \[\sqrt{1+Δx/x}=1+a\] for some small \(a\). We will have \[(1+a)^2=1+Δx/x\]

But \((1+a)^2≈ 1+2a\) when \(a\) is small

\[1+Δx/x = (1+a)^2 ≈ 1+2a\]

Thus \(1+Δx/x≈ 1+2a\) and we find \[a=\frac 1 2 \frac{Δx}{x}\] Therefore \[\sqrt{x±Δx}=\sqrt{x}⋅\sqrt{1+Δx/x} = \sqrt x \left(1 ± \frac 1 2 \frac{Δx}{x}\right)\]

To calculate \((x ± Δx)/(y ± Δy)\)
we do as in multiplication.

We calculate the relative errors and we sum them \[\frac x y \left(1 ± \left(\frac{Δx} x + \frac{Δy}
y \right)\right)\]

Finally we expand the result \[\frac x y ± \frac x y \left(\frac{Δx} x + \frac{Δy} y \right)\]

To understand the formula for the division, we write it as a multiplication \[\frac{x ± Δx}{y ± Δy} = (x ± Δx) × (y ± Δy)^{-1}\] and then we have \[ x\left(1 ± \frac{Δx} x \right) y^{-1}\left(1 ± \frac{Δy} y \right)^{-1} = \frac x y \left(1 ± \frac{Δx} x \right)\left(1 ± \frac{Δy} y \right)^{-1}\]

We need an auxiliary step. This is a small
*parenthesis*.

Let’s assume that \(r\) is a small
number (\(r<<1\)). If \[S = 1+r+r^2+r^3+\cdots\] then we can
multiply everything by \(r\) and we get
\[rS = r+r^2+r^3+\cdots\] Then \(S - rS =1\) and therefore \(S = \frac{1}{1-r} = (1-r)^{-1}\)

\[(1-r)^{-1} = 1+r+r^2+r^3+\cdots\]
Since \(r\) is very small, then \(r^2\) is even much smaller, and so
on.

Therefore \[(1-r)^{-1} ≈ 1+r\]

In the same way \[(1+r)^{-1} = 1-r+r^2-r^3+\cdots ≈ 1-r\]

…in the formula for division \[\begin{aligned} x\left(1 ± \frac{Δx} x \right)y^{-1}\left(1 ± \frac{Δy} y \right)^{-1} &= \frac x y \left(1 ± \frac{Δx} x \right)\left(1 ± \frac{Δy} y \right)^{-1}\\ &= \frac x y \left(1 ± \frac{Δx} x \right)\left(1 ± \frac{Δy} y \right) \end{aligned}\]

Assuming that the errors are small,

we can find the error for any “reasonable” function

For any derivable function \(f\) we have \[f(x±Δx) ≈ f(x) ± \frac{df}{dx}(x)\cdot Δx\] This is proved using calculus (Taylor’s Theorem)

We will study it later

\[\frac{dx^2}{dx}=2x\] therefore \[\begin{aligned} (x ±Δx)^2& = x^2 ± 2x\cdotΔx\\ & = x^2 ± 2x^2\cdot\frac{Δx}{x} \end{aligned}\]

\[\frac{d\sqrt x}{dx}=\frac{1}{2\sqrt x}\] therefore \[\begin{aligned} \sqrt{x ±Δx}& = \sqrt x ± \frac{1}{2\sqrt x}\cdot Δx\\ & = \sqrt x ± \frac{1}{2}\sqrt x\cdot \frac{Δx}{x} \end{aligned}\]

\[ \begin{aligned} (x ± Δx) + (y ± Δy) & = (x+y) ± (Δx+Δy)\\ (x ± Δx) - (y ± Δy) & = (x-y) ± (Δx+Δy)\\ (x ± Δx\%) \times (y ± Δy\%)& =xy ± (Δx\% + Δy\%)\\ (x ± Δx\%) ÷ (y ± Δy\%)& =x/y ± (Δx\% + Δy\%)\end{aligned} \]

Here \(Δx\%\) represents the
*relative* uncertainty, that is \(Δx/x\)

We use absolute uncertainty for + and -, and relative uncertainty for ⨉ and ÷

Calculate the uncertainty in the density of the stone ball