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

420 Chapter 12 Applications of differentiation
Using the Newton--Raphson technique the value ofx 2
isfound:
x 2
=x 1
− f(x 1 )
f ′ (x 1
) =7.5−(−723) = 8.41
796
An improved estimate of the root of e x − 6x 3 = 0 isx = 8.41. To two decimal places
the trueanswer isx = 8.05.
The Newton--Raphson technique can be used repeatedly as illustrated in
Example12.7.Thisgeneratesasequenceofapproximatesolutionswhichmayconverge
tothe required root. Eachapplication ofthe methodisknown asan iteration.
Example12.7 A root of 3sinx = x is near to x = 2.5. Use two iterations of the Newton--Raphson
technique tofind a more accurateapproximation.
Solution Theequation mustfirst bewritten inthe form f (x) = 0,thatis
Then
Then
f(x)=3sinx−x=0
x 1
= 2.5
f(x)=3sinx−x f(x 1
)=−0.705
f ′ (x)=3cosx−1 f ′ (x 1
) = −3.403
x 2
=2.5− (−0.705)
(−3.403) = 2.293
The process isrepeated withx 1
= 2.293 as the initial approximation:
Then
x 1
= 2.293 f (x 1
) = −0.042 f ′ (x 1
) = −2.983
x 2
= 2.293 − (−0.042)
(−2.983) = 2.279
Using two iterations of the Newton--Raphson technique, we obtainx = 2.28 as an improved
estimateof the root.
Example12.8 An approximaterootof
Solution We have
x 3 −2x 2 −5=0
isx = 3. By using the Newton--Raphson technique repeatedly, determine the value of
the rootcorrecttotwo decimalplaces.
Hence
x 1
=3
f(x)=x 3 −2x 2 −5 f(x 1
)=4
f ′ (x)=3x 2 −4x f ′ (x 1
)=15
x 2
=3− 4
15 = 2.733