Errors

Identifying error sources and types is crucial in any measurement.

Types of errors

1. Systematic errors: Result from known causes (e.g., incorrect formulas, miscalibrated devices).
2. Random errors: Result from unpredictable sources (e.g., environmental conditions).

Sources of errors

Errors in calculations and measurements relate to accuracy and precision.

1. Accuracy: How close a computed/measured value is to the actual value.
2. Precision: How close individual computed/measured values are to each other.

Main error sources:

1. Rounding-off errors: Approximations during computations due to the inability to represent true values exactly (e.g., $\pi \approx 3.14$).
2. Truncation errors: Approximations of infinite mathematical processes (e.g., terminating an iterative method or using a finite number of terms in an infinite series).
3. Data uncertainty: Errors in data due to systematic or random errors in measurements.

Example of truncation error

The Maclaurin series for $e^x$ is:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots$

Using the first three terms to approximate $e^x$:

$e^x \approx 1 + x + \frac{x^2}{2!}$

The truncation error is the difference between the actual value and the approximation:

$\text{Truncation Error} = e^x - \left(1 + x + \frac{x^2}{2!}\right) = \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots$

Errors are expressed as absolute or relative errors.

Absolute error ($\epsilon_a$)

Measures how far a measurement is from the actual value:

$\epsilon_a = |x - x^*|$

Where:

• $x$ is the actual/true value.
• $x^*$ is the measured/approximated value.

Relative error ($\epsilon_r$)

The ratio of the absolute error to the absolute value of the actual value:

$\epsilon_r = \frac{|x - x^*|}{|x|}$

Percentage relative error ($\epsilon_p$):

$\epsilon_p = \frac{|x - x^*|}{|x|} \times 100\%$

Difference between absolute and relative error

Feature	Absolute error	Relative error
Definition	Difference between measured and actual value	Ratio of absolute error to actual value
Meaning	How large the error is	How good/bad the error is (relative to the size of the value)
Units	Same units as the measured quantity	Unitless (or percentage)

Numerical computations involve a series of arithmetic operations. Understanding how errors propagate through these operations is crucial.

Effects of absolute errors

Let $\epsilon_1 = x_1 - x_1^*$ and $\epsilon_2 = x_2 - x_2^*$ be the absolute errors in approximating $x_1$ and $x_2$, respectively.

Addition and subtraction

The maximum absolute error in addition or subtraction is:

$|(x_1 \pm x_2) - (x_1^* \pm x_2^*)| = |(x_1 - x_1^*) \pm (x_2 - x_2^*)| \le |\epsilon_1| + |\epsilon_2|$

Thus, the maximum absolute error in adding/subtracting two quantities is at most the sum of their absolute errors.

For $n$ approximate quantities with absolute errors $\epsilon_k$ ($k = 1, 2, ..., n$), the maximum absolute error is:

$\sum_{k=1}^{n} |\epsilon_k| = |\epsilon_1| + |\epsilon_2| + ... + |\epsilon_n|$

Multiplication

The absolute error in the product is:

$|x_1x_2 - x_1^*x_2^*| = |x_1x_2 - x_1^*x_2 + x_1^*x_2 - x_1^*x_2^*| = |(x_1 - x_1^*)x_2 + x_1^*(x_2 - x_2^*)|$

$\le |x_2\epsilon_1| + |x_1^*\epsilon_2| \approx |x_2\epsilon_1| + |x_1\epsilon_2|$

(since $x_1^* \approx x_1$)

Division

The absolute error in the quotient is:

$\left|\frac{x_1}{x_2} - \frac{x_1^*}{x_2^*}\right| = \left|\frac{x_1x_2^* - x_1^*x_2}{x_2x_2^*}\right| = \left|\frac{x_1(x_2+\epsilon_2) - (x_1+\epsilon_1)x_2}{x_2x_2^*}\right| = \left|\frac{x_1\epsilon_2 - x_2\epsilon_1}{x_2x_2^*}\right|$

$\approx \left|\frac{x_1\epsilon_2 - x_2\epsilon_1}{x_2^2}\right| \le \frac{|x_1||\epsilon_2| + |x_2||\epsilon_1|}{|x_2|^2} = \frac{|\epsilon_1|}{|x_2|} + \frac{|x_1||\epsilon_2|}{|x_2|^2}$

Effects of relative errors

Let $\epsilon_{r1} = \frac{x_1 - x_1^*}{x_1}$ and $\epsilon_{r2} = \frac{x_2 - x_2^*}{x_2}$ be the relative errors.

Multiplication

The relative error in the product is:

$\left|\frac{x_1x_2 - x_1^*x_2^*}{x_1x_2}\right| = \left|\frac{x_1x_2 - (x_1-\epsilon_1)(x_2-\epsilon_2)}{x_1x_2}\right| = \left|\frac{x_1x_2 - x_1x_2 + x_1\epsilon_2 + x_2\epsilon_1 - \epsilon_1\epsilon_2}{x_1x_2}\right| = \left|\frac{x_1\epsilon_2 + x_2\epsilon_1 - \epsilon_1\epsilon_2}{x_1x_2}\right|$

$\approx \left|\frac{x_1\epsilon_2 + x_2\epsilon_1}{x_1x_2}\right| = \left|\frac{\epsilon_2}{x_2} + \frac{\epsilon_1}{x_1}\right| = |\epsilon_{r2} + \epsilon_{r1}|$

$\le |\epsilon_{r1}| + |\epsilon_{r2}|$

Thus, the relative error in a product is approximately the sum of the relative errors.

Division

The relative error in the quotient is:

$\left|\frac{\frac{x_1}{x_2} - \frac{x_1^*}{x_2^*}}{\frac{x_1}{x_2}}\right| = \left|\frac{x_1x_2^* - x_1^*x_2}{x_1x_2^*}\right| \cdot \frac{|x_2|}{|x_1|} = \left|\frac{x_1(x_2+\epsilon_2) - (x_1+\epsilon_1)x_2}{x_1(x_2+\epsilon_2)}\right| \frac{|x_2|}{|x_1|} = \left|\frac{x_1\epsilon_2 - x_2\epsilon_1}{x_1x_2 + x_1\epsilon_2}\right|\frac{|x_2|}{|x_1|} \approx \left|\frac{\epsilon_2}{x_2} - \frac{\epsilon_1}{x_1}\right| \le |\epsilon_{r1}| + |\epsilon_{r2}|$

Thus, the relative error in a quotient is approximately the sum of the relative errors.

Find the truncation error in terminating the series of $e^x$ for $x = 0.2$ after the first three terms. Give the answer correct to four decimal places.

Series: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + ...$

Approximation: $e^x \approx 1 + x + \frac{x^2}{2!}$

Truncation error: $\epsilon_t = e^x - \left(1 + x + \frac{x^2}{2!}\right) = \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + ...$

Substituting $x = 0.2$:

$\epsilon_t = \frac{(0.2)^3}{3!} + \frac{(0.2)^4}{4!} + \frac{(0.2)^5}{5!} + ... = \frac{0.008}{6} + \frac{0.0016}{24} + \frac{0.00032}{120} + ...$

$\approx 0.001333 + 0.000067 + 0.0000026 + ...$

$\approx 0.00140266...$

Thus, $\epsilon_t \approx 0.0014$ (correct to four decimal places).

If the exact value is $x = 2.2056 \times 10^{-4}$ and the approximated value is $x^* = 2.1056 \times 10^{-4}$, find the absolute and relative errors.

Absolute error: $\epsilon_a = |x - x^*| = |2.2056 \times 10^{-4} - 2.1056 \times 10^{-4}| = |0.1 \times 10^{-4}| = 1 \times 10^{-5}$

Relative error: $\epsilon_r = \frac{|\epsilon_a|}{|x|} = \frac{1 \times 10^{-5}}{2.2056 \times 10^{-4}} \approx 0.04534$

Percentage relative error: $\epsilon_p = \epsilon_r \times 100\% \approx 0.04534 \times 100\% = 4.534\%$

Given $p = 10.6 \pm 0.2$ cm and $q = 25.5 \pm 0.3$ cm, calculate the maximum possible error in $p + q$.

Sum: $p + q = 10.6 + 25.5 = 36.1$ cm

Maximum error: $0.2 + 0.3 = 0.5$ cm

Result: $36.1 \pm 0.5$ cm

The lengths of two objects are $21.5 \pm 0.1$ cm and $17.8 \pm 0.3$ cm. Find the sum of the lengths with the limit of errors.

Sum of lengths: $(21.5 + 17.8) \pm (0.1 + 0.3) = 39.3 \pm 0.4$ cm

The initial temperature is $36.4 \pm 0.3$°C, and the final temperature is $63.6 \pm 0.4$°C. Find the increase in temperature.

Increase in temperature: $(63.6 - 36.4) \pm (0.4 + 0.3) = 27.2 \pm 0.7$°C

A silver string has mass $m = 10 \pm 0.1$ g, length $l = 50 \pm 0.5$ mm (5 cm), width $w = 10 \pm 0.2$ mm (1 cm), and thickness $h = 2 \pm 0.1$ mm (0.2 cm). Determine the error in the density.

Density: $\rho = \frac{m}{V} = \frac{m}{lwh}$

Using logarithmic differentiation to estimate the error:

$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta l}{l} + \frac{\Delta w}{w} + \frac{\Delta h}{h}$

$\frac{\Delta\rho}{\rho} = \frac{0.1}{10} + \frac{0.5}{50} + \frac{0.2}{10} + \frac{0.1}{2} = 0.01 + 0.01 + 0.02 + 0.05 = 0.09$

$\rho = \frac{10}{5 \times 1 \times 0.2} = 10 \text{ g/cm}^3$

$\Delta\rho = 0.09 \times 10 = 0.9 \text{ g/cm}^3$

The error in density is $0.9$ g/cm³.

Calculate the percentage error in calculating $g$ from $T = 2\pi\sqrt{\frac{l}{g}}$, if $l$ and $T$ have errors of $\pm 3\%$ and $\pm 4\%$, respectively.

Squaring the equation: $T^2 = 4\pi^2\frac{l}{g}$

$g = \frac{4\pi^2l}{T^2}$

Using logarithmic differentiation:

$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$

Percentage error in $g$:

$\frac{\Delta g}{g} \times 100\% = (3\% + 2 \times 4\%) = 3\% + 8\% = 11\%$