Civil Engineering Reference
In-Depth Information
′′
()
()
f x x
∆
()
+
()
′
()
+
n
fx xf
∆
x
=
0
n
n
2
The exact value of Δ
x
cannot
be determined from this equation since only
the first three terms of the infinite series were used in the calculation.
However, a close approximation of the root is a result. When using this
equation to calculate Δ
x
, a quadratic must be solved yielding two possible
roots. In order to avoid this problem, Δ
x =
−
f
(
x
n
)
/f
′
(
x
n
) from Newton's
tangent may be substituted into the bracketed term only.
′′
()()
′
()
=
f xfx
fx
()
+
()
′
()
−
n
n
fx xfx
∆
0
n
n
2
n
Solving for Δ
x
we obtain the following:
()
fx
fx
f xfx
fx
∆
x
=−
n
′′
()()
′
()
′
()
−
n
n
n
2
n
Observing Figure 1.9 we see that Δ
x = x
n+
1
−
x
n
. Substituting into the pre-
vious equation, Equation 1.3 is obtained as follows:
()
fx
fx
f xfx
fx
n
x
+
=−
x
(1.3)
′′
()()
′
()
n
1
n
′
()
−
n
n
n
2
n
If the first derivative is small, the slope is close to zero near the value
and the next approximation may be inaccurate. Therefore, use the second
derivative term as follows:
′
()
=
()
+
()
′′
()
()
fx
0
n
f x x
∆
n
fx x
∆
=
0
n
2
−
()
=
()
′′
()
(()
2
f x x
∆
n
fx x
∆
n
2
+
()
fx
f x
n
∆
x
2
′′
()
=
0
n
2
