Error Propagation

Approximations in measurements introduce errors that carry through mathematical operations during calculations.

When I measure a quantity with an instrument, the result isn’t the exact real value but rather an estimate (or approximation) that includes some error.

This error can stem from two main sources:

• Limitations of the instrument (sensitivity, precision).
• Observational errors or environmental factors.

When I use these approximated values in calculations, the uncertainty in the final result inevitably increases, and I have to take it into account.

In general, using approximated values introduces a margin of error into the calculation.

These errors propagate differently with each operation, depending on the type of mathematical operation involved (addition, multiplication, or exponentiation).

Why is it important to understand error propagation? Knowing how errors propagate helps me gauge the reliability of the final result and estimate its uncertainty, or trustworthiness. Ignoring error propagation can lead to misleading results or incorrect conclusions.

How do errors propagate in different mathematical operations?

According to the error propagation theorem, the way errors propagate depends on the mathematical operation performed.

• Addition and Subtraction

  In addition or subtraction, the absolute error of the result is the sum of the absolute errors of the operands, or the individual measurements. $$ \text{Total Error} = \Delta x + \Delta y $$

  This is because if each measured value can vary within a certain range of uncertainty, then the sum or difference will vary accordingly.
• Multiplication and Division

  For multiplication or division, the relative error of the result is the sum of the relative errors of the operands. $$ \frac{\Delta (xy)}{xy} = \frac{\Delta x}{x} + \frac{\Delta y}{y} $$

  In other words, with multiplication, it’s the relative error — the ratio of absolute error to measured value — that adds up. The same rule applies to division, where relative errors combine.
• Multiplication and Division by a Constant

  When a quantity $ x $ is multiplied by a numerical constant $ k $, the absolute error of the result is given by multiplying the absolute error $ \Delta x $ of the quantity by the constant $ k $. $$ \Delta kx = k \cdot \Delta x $$

  For division, the absolute error of the result is calculated by dividing the absolute error of the quantity by the constant $ k $. $$ \Delta ( \frac{x}{k} ) = \frac{\Delta x}{k} $$
• Exponents

  When raising a measurement to a power \( n \), the relative error is multiplied by \( n \).

  For example, if calculating the volume of a cube, each side has a relative uncertainty that multiplies by 3 because there are three sides in the volume calculation, corresponding to an exponent of 3.

So, each mathematical operation on approximated values brings along its own degree of uncertainty in a unique way.

Understanding how errors propagate allows me to interpret and use measurement results rigorously.

A Practical Example

A] Addition

I need to measure the dimensions of a rectangular table using a tape measure with a sensitivity of up to 1 millimeter.

The table’s measurements are:

• \( X = (100.0 \pm 0.1) \ \text{cm} \)
• \( Y = (50.0 \pm 0.1) \ \text{cm} \)

The instrument’s sensitivity introduces an uncertainty range of 1 millimeter, or 0.1 centimeters, in each measurement.

Using this data, I analyze the error propagation in various mathematical operations.

The semi-perimeter \( p \) of the rectangle is:

$$ p = X + Y = 100.0 + 50.0 = 150.0 \ \text{cm} $$

According to the error propagation theorem, the absolute error in the sum is the sum of the absolute errors of the terms

$$ E = \Delta X + \Delta Y = 0.1 + 0.1 = 0.2 \ \text{cm} $$

So, I can express the semi-perimeter as:

$$ p = (150.0 \pm E) = (150.0 \pm 0.2) \ \text{cm} $$

B] Multiplication

Continuing with the table example, with the following measurements:

• \( X = (100.0 \pm 0.1) \ \text{cm} \)
• \( Y = (50.0 \pm 0.1) \ \text{cm} \)

The area \( a \) of the table is given by the product \( X \times Y \)

$$ a = XY = 100.0 \times 50.0 = 5000.0 \ \text{cm}^2 $$

According to the propagation theorem, the relative error in multiplication is the sum of the relative errors

$$ \frac{\Delta (XY) }{XY} = \frac{\Delta X}{X} + \frac{\Delta Y }{Y} $$

$$ \frac{\Delta (XY) }{XY} = \frac{0.1}{100} + \frac{0.1}{50} $$

$$ \frac{\Delta (XY) }{XY} = 0.003 $$

So, the absolute error in the product is

$$ \Delta (XY) = 0.003 \cdot XY $$

$$ \Delta (XY) = 0.003 \cdot 5000.0 \ \text{cm}^2 $$

$$ \Delta (XY) = 15.0 \ \text{cm}^2 $$

Thus, the area of the table can be expressed as:

$$ a = XY \pm \Delta XY = 5000.0 \pm 15.0 \ \text{cm}^2 $$

C] Exponents

Let’s see how error propagation works when calculating a power applied to a measurement.

For example, I want to calculate the volume \( V \) of a rectangular box with the following measurements and associated absolute errors:

• \( X = 50.0 \pm 0.1 \ \text{cm} \)
• \( Y = 50.0 \pm 0.1 \ \text{cm} \)
• \( Z = 50.0 \pm 0.1 \ \text{cm} \)

The volume \( V \) of the box is given by the product of its three dimensions:

$$ V = X \cdot Y \cdot Z $$

Calculating the volume using nominal values (without errors):

$$ V = 50.0 \ \text{cm} \cdot 50.0 \ \text{cm} \cdot 50.0 \ \text{cm} = 125000 \ \text{cm}^3 $$

Knowing that the relative error of the volume is the sum of the relative errors of each dimension:

$$ \frac{ \Delta V }{ V } = \left( \frac{\Delta X}{X} + \frac{\Delta Y}{Y} + \frac{\Delta Z}{Z} \right) $$

Then, the absolute error in the volume is:

$$ \Delta V = V \left( \frac{\Delta X}{X} + \frac{\Delta Y}{Y} + \frac{\Delta Z}{Z} \right) $$

Now I calculate the relative error for each dimension \( X \), \( Y \), and \( Z \):

• Relative error of \( X \) $$ \frac{\Delta X}{X} = \frac{0.1}{50.0} = 0.002 $$
• Relative error of \( Y \) $$ \frac{\Delta Y}{Y} = \frac{0.1}{50.0} = 0.002 $$
• Relative error of \( Z \) $$ \frac{\Delta Z}{Z} = \frac{0.1}{50.0} = 0.002 $$

Adding up these relative errors gives the total relative error in the volume:

$$ \frac{\Delta V}{V} \approx 0.002 + 0.002 + 0.002 = 0.006 $$

To find the absolute error \( \Delta V \), I multiply the total relative error by the nominal volume \( V \):

$$ \Delta V = V \cdot 0.006 $$

$$ \Delta V = 125000 \cdot 0.006 = 750 \ \text{cm}^3 $$

Thus, the volume of the box, considering measurement errors, can be expressed as:

$$ V = 125000 \pm 750 \ \text{cm}^3 $$

Proof

A] Addition and Subtraction

I need to prove that the absolute error in the sum of two measured quantities is the sum of the absolute errors of the individual terms.

I have two approximate measurements:

• \( x \pm \Delta x \), where \( x \) is the measured value, and \( \Delta x \) is the absolute error associated with \( x \).
• \( y \pm \Delta y \), where \( y \) is the measured value, and \( \Delta y \) is the absolute error associated with \( y \).

I want to calculate the sum of the two measurements, including their errors.

When adding the two measurements with their respective errors, I get:

$$ (x \pm \Delta x) + (y \pm \Delta y) $$

Therefore, the sum can take two extreme values:

• Maximum case: \( (x + \Delta x) + (y + \Delta y) = (x + y) + \Delta x + \Delta y \)
• Minimum case: \( (x - \Delta x) + (y - \Delta y) = (x + y) - \Delta x - \Delta y \)

This means the range for the result of \( x + y \) is given by:

$$ (x + y) \pm (\Delta x + \Delta y) $$

So, I have shown that the sum of the measurements is \( x + y \), and its absolute error is \( \Delta x + \Delta y \).

Thus:

$$ (x \pm \Delta x) + (y \pm \Delta y) = (x + y) \pm (\Delta x + \Delta y) $$

In other words, the absolute error in addition is the sum of the absolute errors of the individual terms.

B] Multiplication and Division

To prove that the relative error in the multiplication of two quantities is the sum of the relative errors of the individual terms, I consider the following measured quantities with their respective absolute errors:

• \( x \pm \Delta x \), where \( x \) is the measured value, and \( \Delta x \) is the absolute error on \( x \).
• \( y \pm \Delta y \), where \( y \) is the measured value, and \( \Delta y \) is the absolute error on \( y \).

I want to calculate the relative error of the product \( x \cdot y \) in terms of the relative errors of \( x \) and \( y \).

The product \( xy \) with its extreme errors can be expressed as:

$$ (x \pm \Delta x) \cdot ( y \pm \Delta y) $$

The product \( xy \) can take the following extreme values:

• Maximum case: \((x + \Delta x)(y + \Delta y)\)
• Minimum case: \((x - \Delta x)(y - \Delta y)\)

I’m interested in expanding the expression for the maximum case, as it represents the positive extreme, to see how the product varies with respect to the error.

Expanding the expression \((x + \Delta x)(y + \Delta y)\):

$$ (x + \Delta x)(y + \Delta y) = xy + x \Delta y + y \Delta x + \Delta x \Delta y $$

Since errors \( \Delta x \) and \( \Delta y \) are typically small compared to \( x \) and \( y \), I can ignore the term \( \Delta x \Delta y \) and approximate:

$$ (x + \Delta x)(y + \Delta y) \approx xy + x \Delta y + y \Delta x $$

Thus, the absolute error in the product \( xy \) is approximately:

$$ \Delta(xy) \approx x \Delta y + y \Delta x $$

The relative error is given by the ratio of the absolute error to the measured value.

$$ \frac{\Delta(xy)}{xy} \approx \frac{x \Delta y + y \Delta x}{xy} $$

I break down this fraction:

$$ \frac{\Delta(xy)}{xy} \approx \frac{x \Delta y}{xy} + \frac{y \Delta x}{xy} $$

$$ \frac{\Delta y}{y} + \frac{\Delta x}{x} $$

Thus, I have shown that the relative error in the product \( xy \) is the sum of the relative errors of the individual terms \( x \) and \( y \):

$$ \frac{\Delta(xy)}{xy} \approx \frac{\Delta x}{x} + \frac{\Delta y}{y} $$

In other words, the relative error in multiplication is the sum of the relative errors of the operands.

And so on.