# It is the mark of an instructed mind to rest assured with that degree of precision that the nature of the subject admits, and not to seek exactness when only an approximation of the truth is possible

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

# Combining uncertainty

## Alternative representation of intervals

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)

## We will assume that $$Δx$$ is smaller than $$x$$

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

## 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 $(x ± Δx) = x(1 ± Δx/x)\\ (y ± Δy) = y(1 ± Δy/y)$

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

Finally we expand the result $xy ± xy(Δx/x + Δy/y)$

## Proof

\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

## 𝑥 squared

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

## Powers of 𝑥

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 $(1 + Δx/x)^n ≈ {n\choose 0} (Δx/x)^0 + {n\choose 1} (Δx/x)^1 = 1 + n(Δx/x)$

## Therefore

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 $(x ± Δx)^n =\left(x\left(1 ± \frac{Δx}{x}\right)\right)^n ≈ x^n\left(1 ± n\frac{Δx}{x}\right)$

## Square root of 𝑥

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

## Since 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 $\sqrt{x±Δx}=\sqrt{x}⋅\sqrt{1+Δx/x} = \sqrt x \left(1 ± \frac 1 2 \frac{Δx}{x}\right)$

## Division

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 $\frac x y ± \frac x y \left(\frac{Δx} x + \frac{Δy} y \right)$

## Why this is true?

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 $x\left(1 ± \frac{Δx} x \right) y^{-1}\left(1 ± \frac{Δy} y \right)^{-1} = \frac x y \left(1 ± \frac{Δx} x \right)\left(1 ± \frac{Δy} y \right)^{-1}$

## To finish we need another formula

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 $$S = \frac{1}{1-r} = (1-r)^{-1}$$

## In other words

$(1-r)^{-1} = 1+r+r^2+r^3+\cdots$ 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 $(1+r)^{-1} = 1-r+r^2-r^3+\cdots ≈ 1-r$

## Putting it back…

…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}

## General formula

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

## Example 1

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

## Example 2

$\frac{d\sqrt x}{dx}=\frac{1}{2\sqrt x}$ 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}

## Summary of error propagation rules

\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 ÷

## Exercise

Calculate the uncertainty in the density of the stone ball