Lab 4: Error Lab · Numerical Methods

Choose an expression and move x. The red curve is the naive formula, the green curve the rewritten one. The vertical axis is the relative error on a log scale: 1e−16 is perfect, 1e0 means every digit is wrong.

Error of the forward and central difference approximations of f′(x) as the step h changes (log scales). Type your own f(x).

Try this

Run it. Which approximation of π is best?
Add 3.14159265 to the list. How many digits does the program report?
Change x_true to math.sqrt(2) and the list to [1.4, 1.41, 1.414, 1.4142].

e⁻²⁰ by series

Try this

Run it. How wrong is the naive series?
Try x = -5 and x = -10. When does the naive sum become useless?
Print the largest term for x = -30. Compare its rounding error (× 1.1e-16) with e^-30.

Variance

Try this

Run it. Which variance is right?
Change 1e8 to 1e9. What happens to the one-pass result?
Change 1e8 to 0. Do both methods now agree?

Q1 Find the absolute, relative and percentage errors if

2/3

is approximated by

0.6667

.

E_a = |0.666667 - 0.6667| = 3.33\times10^{-5}

;

E_r = 3.33\times10^{-5}/0.666667 = 5.0\times10^{-5}

;

E_p = 0.005\%

.

Q2 If

u = \dfrac{5xy^2}{z^3}

and

\Delta x = \Delta y = \Delta z = 0.001

at

x = y = z = 1

, find the maximum relative error in

u

.

\Delta u/u = \Delta x/x + 2\Delta y/y + 3\Delta z/z = (1+2+3)(0.001) = 0.006

.

Q3 The sides of a rectangle are

l = 12.0 \pm 0.1

cm and

w = 8.0 \pm 0.1

cm. Find the area and its maximum error.

A = 96

cm².

\Delta A \approx w\Delta l + l\Delta w = 0.8 + 1.2 = 2.0

cm² (relative 2.08%).

Q4 Distinguish round-off error and truncation error with one example each.

Round-off: from finite digits, e.g.

1/3 \to 0.3333

. Truncation: from stopping an infinite process, e.g.

e^x \approx 1 + x + x^2/2

; the neglected terms are the truncation error.

Q5 Evaluate

\sqrt{9.01} - 3

in 4-significant-digit arithmetic, then by a rewritten formula. Compare with the true value

0.0016662

.

Naive:

\sqrt{9.01} = 3.002

, so

3.002 - 3 = 0.002

(20% error). Rewritten:

0.01/(\sqrt{9.01} + 3) = 0.01/6.002 = 0.001666

(0.01% error).

Q6 Why does

1 - \cos x

give 0 in double precision for

x = 10^{-9}

? Give an accurate formula.

\cos(10^{-9}) = 1 - 5\times10^{-19}

, which rounds to exactly 1.0 (the gap below 1 is

1.1\times10^{-16}

). Use

2\sin^2(x/2)

, which gives

5\times10^{-19}

.