082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

12.4 The Newton--Raphson method for solving equations 419
f (x)
y = f (x)
B
^
x
C A
x 2 x 1 x
Figure12.14
Thetangent at Bintersects thexaxisat C.
Suppose we wish to find a root of f (x) = 0. Figure 12.14 illustrates the curvey =
f (x). Roots of the equation f (x) = 0 correspond to where the curve cuts the x axis.
One such root is illustrated and is labelledx = ˆx. Suppose we know thatx = x 1
is an
approximate solution. Let A be the point on thexaxis wherex = x 1
and let B be the
pointonthecurvewherex =x 1
.ThetangentatBisdrawnandcutsthexaxisatCwhere
x =x 2
.Clearlyx =x 2
isabetterapproximationto ˆxthanx 1
.Wenowfindx 2
intermsof
the known value,x 1
.
Hence,
AB = distance of Babove thexaxis = f (x 1
)
CA=x 1
−x 2
gradient of lineCB = AB
CA = f(x 1 )
x 1
−x 2
ButCB isatangent tothe curve atx =x 1
and so has gradient f ′ (x 1
). Hence,
f ′ (x 1
) = f(x 1 )
x 1
−x 2
x 1
−x 2
= f(x 1 )
f ′ (x 1
)
and therefore,
x 2
=x 1
− f(x 1 )
f ′ (x 1
)
(12.7)
Equation (12.7) is known as the Newton--Raphson formula. Knowing an approximate
root of f (x) = 0, that is x 1
, the Newton--Raphson formula enables us to calculate an
improved approximate root,x 2
.
Example12.6 Given thatx 1
= 7.5 is an approximate root of e x −6x 3 = 0, use the Newton--Raphson
techniquetofind animproved value.
Solution x 1
=7.5
f(x)=e x −6x 3 f(x 1
)=−723
f ′ (x)=e x −18x 2 f ′ (x 1
) =796