Class 5.1: Error propagation

Methodology of Scientific Research

Andrés Aravena, PhD

March 17, 2022

The aim of science is not to open the door to infinite wisdom, but to set a limit to infinite error

Bertolt Brecht, Life of Galileo (1939)

Measurements have uncertainty

Uncertainty of measurement is the doubt about the result of a measurement, due to

resolution
random errors
systematic errors

Uncertainty must be declared

How big is the margin? How bad is the doubt?

We declare an interval: [x_min, x_max]

Most of the time we write x ± 𝚫x

Example: 20cm ± 1cm

Error versus uncertainty

Do not to confuse error and uncertainty

Error is the difference between the measured and the “true” value

Uncertainty is a quantification of the doubt about the result

Whenever possible we try to correct for any known errors

But any error whose value we do not know is a source of uncertainty

Combined uncertainty

Uncertainty of a single read

For a single read, the uncertainty depends at least on the instrument resolution

For example, my old water heater showed temperature with 5°C resolution: 50, 55, 60,…

Real vs measured value

If it shows 55°C, the real temperature is somewhere between 53°C and 57°C

We write 55°C ± 2.5°C

For a single read, 𝚫x = half of the resolution

Combining Uncertainty

Sum of two measurements:
\[(x ± 𝚫x) + (y ± 𝚫y) = (x+y) ± (𝚫x+𝚫y)\]

Difference between measurements:
\[(x ± 𝚫x) - (y ± 𝚫y) = (x-y) ± (𝚫x+𝚫y)\]

Multiplying Uncertainty

To calculate \((x ± 𝚫x) × (y ± 𝚫y)\) we first write the uncertainty as percentage

\[(x ± 𝚫x/x\%) × (y ± 𝚫y/y\%)\]

Then we sum the percentages:
\[xy ± (𝚫x/x + 𝚫y/y)\%\]

Finally we convert back to the original units:
\[xy ± xy(𝚫x/x + 𝚫y/y)\]

Proof

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

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

The easy rule for error propagation

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

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

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

General formula

Assuming that the errors are small compared to the main value, we can find the error for any “reasonable” function

Taylor’s Theorem says that, for any derivable function \(f,\) we have \[f(x±𝚫x) = f(x) ± \frac{df}{dx}(x)\cdot 𝚫x + \frac{d^2f}{dx^2}(x+\varepsilon)\cdot \frac{𝚫x^2}{2}\] When \(𝚫x\) is small, we can ignore the last part.

Example 1

\[\begin{align} (x ±𝚫x)^2& = x^2 ± 2x\cdot𝚫x\\ & = x^2 ± 2x^2\cdot\frac{𝚫x}{x} \\ & = x^2 ± 2𝚫x\% \end{align}\]

Example 2

\[\begin{align} \sqrt{x ±𝚫x}& = \sqrt x ± \frac{1}{2\sqrt x}\cdot 𝚫x\\ & = \sqrt x ± \frac{1}{2}\sqrt x\cdot \frac{𝚫x}{x}\\ & = \sqrt x ± \frac{1}{2}𝚫x\% \end{align}\]