CS205 Homework #5 Solutions
Problem 1
[Heath 5.5, p.248]
1. Show that the iterative method
x
k
+1
=
x
k

1
f
(
x
k
)

x
k
f
(
x
k

1
)
f
(
x
k
)

f
(
x
k

1
)
is mathematically equivalent to the secant method for solving a scalar nonlinear equa
tion
f
(
x
) = 0.
2. When implemented in ﬁniteprecision ﬂoatingpoint arithmetic, what advantages or
disadvantages does the formula given in part (1) have compared with the formula for
the secant method (given in the notes and in Heath, section 5.5.4)?
Solutions
1. Starting with the secant method update formula is given as
x
k
+1
=
x
k

f
(
x
k
)(
x
k

x
k

1
)
f
(
x
k
)

f
(
x
k

1
)
=
x
k
(
f
(
x
k
)

f
(
x
k

1
))

f
(
x
k
)(
x
k

x
k

1
)
f
(
x
k
)

f
(
x
k

1
)
=
x
k
f
(
x
k
)

x
k
f
(
x
k

1
)

x
k
f
(
x
k
) +
x
k

1
f
(
x
k
)
f
(
x
k
)

f
(
x
k

1
)
=
x
k

1
f
(
x
k
)

x
k
f
(
x
k

1
)
f
(
x
k
)

f
(
x
k

1
)
2. When we’re near the solution
x
k

1
and
x
k
are close to each other and thus their
diﬀerence is near zero. Similarly
f
(
x
k

1
) and
f
(
x
k
) are close to each other so their
diﬀerence is near zero. Thus we have an indeﬁnite form in secant computing (
x
k

x
k

1
)
/
(
f
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 Fall '07
 Fedkiw
 Numerical Analysis, xk, Secant method, Rootfinding algorithm

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