SME
Numerical Methods for Engineers · Problem type
Secant method: one step
Use this for “Take one secant step for x³ − x − 3 = 0 from x₀ = 1.5 and x₁ = 2.”
Diagram
The secant through (x₀, f(x₀)) and (x₁, f(x₁)) has slope s = 8.25.
It
meets
the
axis
at
x₂
=
x₁
−
f
(x₁)
s
=
1.636,
the
next
guess;
εₐ
=
22.22%.
This is an example. Type your own values to start a new problem.
f(x)
=
c₃
x³
+
c₂
x²
+
(
c₁
)
x
+
(
c₀
)
x₂
=
x₁
−
f(
)
s
=
x₀
First guess
example
f(x₀)
Function value at x₀
worked out (not typed)
f(x₁)
Function value at x₁
worked out (not typed)
εₐ
Approximate error
calculated · %
Formulas
f
(
x₀
)
=
c₃
×
x₀
3
+
c₂
×
x₀
2
+
c₁
×
x₀
+
c₀
−1.125
=
1
×
1.5
3
+
0
×
1.5
2
+
(−1)
×
1.5
+
(−3)
f
(
x₁
)
=
c₃
×
x₁
3
+
c₂
×
x₁
2
+
c₁
×
x₁
+
c₀
3
=
1
×
2
3
+
0
×
2
2
+
(−1)
×
2
+
(−3)
s
=
f
(
x₁
) −
f
(
x₀
)
x₁
−
x₀
8.25
=
3 − (−1.125)
2 − 1.5
x₂
=
x₁
−
f
(
x₁
)
s
1.6364
=
2
−
3
8.25
εₐ
=
|
x₂
−
x₁
|
÷
|
x₂
|
×
100
22.2222 = |1.6364 − 2| ÷ |1.6364| × 100
Assumptions
•
f(x) = c₃x³ + c₂x² + c₁x + c₀, known only by its values: no derivative is used.
•
The line through (x₀, f(x₀)) and (x₁, f(x₁)) stands in for the tangent: x₂ = x₁ − f(x₁)(x₁ − x₀) ÷ (f(x₁) − f(x₀)).
•
The two starting points need not bracket the root, so the method may run away.
Step-by-step
WE KNOW
c₃ = 1, c₂ = 0, c₁ = −1, c₀ = −3, x₀ = 1.5, x₁ = 2
FIND
function value at x₀ (f(x₀)), function value at x₁ (f(x₁)), secant slope (s), next guess (x₂), approximate error (εₐ)
Step 1 · Find function value at x₀ (f(x₀))
Use f(x₀) = c₃ × x₀³ + c₂ × x₀² + c₁ × x₀ + c₀
Put the first guess into f.
f
(
x₀
)
=
1
×
1.5
3
+
0
×
1.5
2
+
(−1)
×
1.5
+
(−3)
f
(
x₀
)
=
1
×
3.375
+
0
×
2.25
−
1
×
1.5
−
3
f
(
x₀
)
=
3.375
−
1.5
−
3
f
(
x₀
)
=
−1.125
Step 2 · Find function value at x₁ (f(x₁))
Use f(x₁) = c₃ × x₁³ + c₂ × x₁² + c₁ × x₁ + c₀
Put the second guess into f.
f
(
x₁
)
=
1
×
2
3
+
0
×
2
2
+
(−1)
×
2
+
(−3)
f
(
x₁
)
=
1
×
8
+
0
×
4
−
1
×
2
−
3
f
(
x₁
)
=
8
−
2
−
3
f
(
x₁
)
=
3
Step 3 · Find secant slope (s)
Use s = (f(x₁) − f(x₀)) ÷ (x₁ − x₀)
Rise over run between the two points: the slope of the secant line, standing in for f′.
s
=
3 − (−1.125)
2 − 1.5
s
=
4.125
0.5
s
=
8.25
Step 4 · Find next guess (x₂)
Use x₂ = x₁ − f(x₁) ÷ s
Follow
the
secant
line
from
(
x₁
,
f
(
x₁
))
down
to
the
axis,
as
Newton
follows
the
tangent.
x₂
=
2
−
3
8.25
x₂
=
2
−
0.3636
x₂
=
1.636364
Step 5 · Find approximate error (εₐ)
Use εₐ = |x₂ − x₁| ÷ |x₂| × 100
The
size
of
the
step
as
a
percent
of
the
new
guess;
the
next
step
uses
x₁
and
x₂
.
εₐ = |1.636364 − 2| ÷ |1.636364| × 100
εₐ = |−0.3636| ÷ |1.6364| × 100
εₐ
=
0.3636
1.636
×
100
εₐ = 0.2222 × 100
εₐ = 22.2222%
Check
Put the numbers back in. Both sides should match.
−1.125
=
1
×
1.5
3
+
0
×
1.5
2
+
(−1)
×
1.5
+
(−3)
✓
3
=
1
×
2
3
+
0
×
2
2
+
(−1)
×
2
+
(−3)
✓
8.25
=
3 − (−1.125)
2 − 1.5
✓
1.636364
=
2
−
3
8.25
✓
22.2222 = |1.636364 − 2| ÷ |1.636364| × 100 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.