Solution of Algebraic and Transcendental Equations5-13
0
10
0
()
()
fx
xx
fx
⇒=−
′
(5.11)
From this we write the Newton–Raphson
iteration formula
1
()
()
n
nn
n
fx
xx
fx
+
=−
′
(5.12)
This formula can also be obtained using Taylor
series expansion of f at x=x
0
:
2
0
0000
()
()()()()()
2!
xx
fxfxxxfxfx
−
=+−++
′′′
(5.13)
For a first approximation x
1
of x we truncate the
series and obtain a linear equation
10100
0
10
0
()()()()0
()
()
fxfxxxfx
fx
xx
fx
=+−=
′
⇒=−
′
(5.14)
From this we obtain iteration formula (5.12).
Newton–Raphson method for solving
equation f(x) = 0
Algorithm NEWTON ( f, f¢, x
0
, Œ, N)
This algorithm computes a solution of f ( x) = 0 given
an initial ...
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