Errors and Floating-Point Arithmetic - Introduction to Scientific Computing

Absolute and Relative Error¶

Before we can analyze floating-point arithmetic, we need precise definitions of error.

Why Relative Error Matters¶

Example 1 (Same Absolute Error, Different Quality)

	$x = 1$ , $x^* = 2$	$x = 10^6$ , $x^* = 10^6 + 1$
Absolute error	1	1
Relative error	$100\%$	$0.0001\%$
Quality	Terrible	Excellent

Both have the same absolute error, but the first is off by 100% while the second is nearly perfect. Relative error captures what matters.

Floating-Point Numbers as Approximations¶

We need to represent all real numbers — from Avogadro’s number ( $6.02 \times 10^{23}$ ) to Planck’s constant ( $6.63 \times 10^{-34}$ ) — using finite resources. The solution is floating-point arithmetic: a finite set of numbers $\mathbb{F}$ that approximates the real line $\mathbb{R}$ , with a guaranteed bound on the relative error of the approximation.

The key fact is this: for any real number $x$ , its floating-point representation $\text{fl}(x)$ satisfies

\text{fl}(x) = x(1 + \varepsilon), \quad |\varepsilon| \leq \mu

(3)

where $\mu$ is machine epsilon (or unit roundoff). This says that floating-point representation introduces a small relative error — the same relative accuracy whether $x$ is tiny or huge.

Definition 2 (Machine Epsilon)

The machine epsilon $\mu$ is the smallest number such that

\frac{|\text{fl}(x) - x|}{|x|} \leq \mu \quad \text{for all } x \in \mathbb{R}

(4)

For a floating-point system with $t$ mantissa bits:

\mu = \frac{1}{2} \times 2^{-t}

(5)

Precision	Mantissa bits	Machine epsilon	Decimal digits
Single (float32)	23	$\approx 5.96 \times 10^{-8}$	~7
Double (float64)	52	$\approx 1.11 \times 10^{-16}$	~16

Solving the Mystery: The Finite Difference Trade-off¶

In the previous chapter, we observed that finite difference errors increase for very small step sizes. Now we have the tools to explain why.

The Computation on a Machine¶

On a computer, function values carry round-off error. By the floating-point guarantee (3), the computed values satisfy

\widetilde{f}(x+h) = f(x+h)(1 + \varepsilon_1), \quad \widetilde{f}(x) = f(x)(1 + \varepsilon_2), \qquad |\varepsilon_i| \leq \mu.

(6)

So the computed forward difference is

\widetilde{D}_h f(x) \;:=\; \frac{\widetilde{f}(x+h) - \widetilde{f}(x)}{h}.

(7)

Separating the Two Errors¶

Expanding:

\widetilde{D}_h f(x) = \underbrace{\frac{f(x+h) - f(x)}{h}}_{\text{exact finite difference}} \;+\; \underbrace{\frac{f(x+h)\,\varepsilon_1 - f(x)\,\varepsilon_2}{h}}_{\text{round-off perturbation}}.

(8)

Applying Taylor’s theorem to the first term:

\frac{f(x+h) - f(x)}{h} = f'(x) + \frac{h}{2}f''(\xi),

(9)

the total error is

\widetilde{D}_h f(x) - f'(x) = \underbrace{\frac{f(x+h)\,\varepsilon_1 - f(x)\,\varepsilon_2}{h}}_{\text{round-off error}} \;+\; \underbrace{\frac{h}{2}f''(\xi)}_{\text{truncation error}}.

(10)

Since $|\varepsilon_i| \leq \mu$ and $f(x+h) \approx f(x)$ for small $h$ :

\left|\widetilde{D}_h f(x) - f'(x)\right| \;\leq\; \underbrace{\frac{2\mu\,|f(x)|}{h}}_{\text{round-off}} \;+\; \underbrace{\frac{h}{2}\,|f''(\xi)|}_{\text{truncation}}

(11)

The Optimal Step Size¶

Minimizing (11) by setting $dE/dh = 0$ :

-\frac{2\mu|f(x)|}{h^2} + \frac{|f''(\xi)|}{2} = 0 \quad \implies \quad h_{\text{opt}} = 2\sqrt{\frac{\mu|f(x)|}{|f''(\xi)|}} \approx \sqrt{\mu}

(12)

where the last approximation assumes $|f(x)| \approx |f''(\xi)|$ .

Remark 3 (Optimal Step Sizes)

Precision	Machine epsilon $\mu$	Optimal $h$ (forward diff)
Single	$\sim 10^{-8}$	$\sim 10^{-4}$
Double	$\sim 10^{-16}$	$\sim 10^{-8}$