Lab 6: Bisection Lab · Numerical Methods

Step through bisection. The band is the current bracket; the table is the iteration table you write by hand.

Type any equation f(x) = 0 in ordinary textbook notation (e.g. x^3 - 2x - 5, 3x + sin x - e^x, x log10(x) - 1.2), choose a method and press Run to tolerance. Use it to check your homework tables, not to replace working them out.

Bisection

Try this

Run it. How many iterations for tol = 1e-4? Does it match log₂(1/tol)?
Change f to e^x − 3x (use math.exp or exp) with [0, 1].
Set tol = 1e-12. How many iterations now?
Try [3, 4]. What does the program print?

The textbook program

Try this

Run it. What root does it print?
Change the stopping test to fabs(f(c)) < 1e-6. Is the answer the same?
Rewrite it with a function double f(double x) instead of #define.

Iteration count

Try this

Run it. Explain the last line (4 decimal places).

Q1 Find a root of

x^3 - 2x - 5 = 0

by bisection, correct to 2 decimal places.

Sign change:

f(2) = -1

,

f(3) = 16

. Midpoints: 2.5 (+5.625), 2.25 (+1.891), 2.125 (+0.346), 2.0625 (−0.351), 2.09375 (−0.009), 2.109375 (+0.167), 2.1015625 (+0.079). Root ≈ 2.09 (true 2.0945515).

Q2 Show that

\cos x = x

has a root in

[0, 1]

and find it by 5 bisection steps.

f(x) = \cos x - x

:

f(0) = 1 > 0

,

f(1) = -0.4597 < 0

. Midpoints 0.5, 0.75, 0.625, 0.6875, 0.71875 → bracket [0.71875, 0.75]; root ≈ 0.734 (true 0.739085).

Q3 How many bisection steps are needed on

[2, 3]

for an error below

10^{-5}

?

n > \log_2(1/10^{-5}) = 16.6

, so 17 steps.

Q4 State two advantages and two disadvantages of the bisection method.

Advantages: always converges with a valid bracket; known error bound and iteration count. Disadvantages: slow (linear); needs a sign change, so it cannot find double roots and can be fooled by discontinuities.

Q5 Why can bisection not find the root of

f(x) = (x - 2)^2

?

f(x) \ge 0

for all

x

: there is no interval with

f(a)f(b) < 0

, so no starting bracket exists.

Q6 Write a C program for bisection on

x^3 + x - 1 = 0

with

a = 0.6

,

b = 0.8

.

See the "textbook program" in the Code tab: #define f(x), sign check, do-while loop with c = (a+b)/2, update a or b, stop when |b − a| ≤ tol, print the root (0.682327).