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Lecture slide for Computational Method.It is more related to maths solving exercise.
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January 22,2015 Thursday Repl class
Dr Azizan 1
FCM2043Computational Methods
Week 2(4), Lecture 6-Finite Difference Approximations
of Higher Derivatives-Lecturer: Dr Azizan
January 22,2015 Thursday Repl class
Dr Azizan 2
Lesson outcome
• At the end of this session, you should be able to use a centered difference approximation of O(h2) to estimate the second derivative of a function.
January 22,2015 Thursday Repl class
Dr Azizan 3
);19(fromitgsubtractinand2by)12(gmultiplyinand
)12......(..........!3
)(!2
)()(')()(
expansion;seriesTaylorForwardtheRecall
)19.....(!3
)2(!2
)2)(()2)((')()(
:)(oftermsin)(forexpansionseriesTaylorforwardawritewethis,doTos.derivativehigherofestimation
numericalderivetousedbecanexpansionseriesTaylor
3)3(
21
3)3(2
2
2
hxfhxfxfxfxf
hfhxfhxfxfxf
xfxf
iiiii
iiii
ii
January 22,2015 Thursday Repl class
Dr Azizan 4
....12
)(7)()(
)20....().........()()()(2)(
)()()(2)()()()(2)(
)()()(2)(;derivativesecondtheafterseriesthetruncateNow
...12
)(7)()()()(2)(
2.....!3
)(2!2
)(2)('2)(2)(2
)19(......!3
)2(!2
)2)(()2)((')()(
2)3(
212
2212
2212
22
12
4)4(
3)3(212
3)3(
21
3)3(2
2
hxfhxfhOwhere
xfhOh
xfxfxf
xfh
RxfxfxfhxfRxfxfxf
Rhxfxfxfxf
Rhxfhxfhxfxfxfxf
Rhxfhxfhxfxfxf
Rhfhxfhxfxfxf
ii
iiii
iiii
iiii
iiii
ni
iiiii
nii
iii
ni
iii
'second forward finite difference'
January 22,2015 Thursday Repl class
Dr Azizan 5
Exercise• Perform manipulations to obtain a 'second
backward finite difference'
......12
)(7)()(where
)21)......(()()(2)()(
2)4(
)3(
221
hxfhxfhO
hOh
xfxfxfxf
ii
iiii
January 22,2015 Thursday Repl class
Dr Azizan 6
Exercise• Perform manipulations to obtain a 'second
centered finite difference'
)23.....(
)()()()(
)(
asexpressedbecandifferencefinitecenteredsecondtheely,Alternativ
......360
)(12
)()(where
)21)......(()()(2)()(
11
4)6(
2)4(
2
22
11
hh
xfxfh
xfxf
xf
hxfhxfhO
hOh
xfxfxfxf
iiii
i
ii
iiii
January 22,2015 Thursday Repl class
Dr Azizan 7
Example
expansion.seriesTaylortheoftermremaindertheofbasisthe
onresultsyourInterpret.derivativesecondtheofvaluetruethewithestimatesyourCompare
.125.0and25.0sizesstepusing2at887625)(
functiontheofderivativesecondtheestimateto)(ofionapproximatdifferencecenteredaUse
23
2
hxxxxxf
hO
January 22,2015 Thursday Repl class
Dr Azizan 8
Solution
28812)2(150)2(is2atderivativesecondtheofvaluetrueThe
12150)(71275)('
887625)(2
23
fx
xxfxxxf
xxxxf
January 22,2015 Thursday Repl class
Dr Azizan 9
Solution
.288)25.0(
)75.1()2(2)25.2()2(
)()(2)()(
)25.2()(25.2102)2()(2
859383975175125.0Using
2
211
11
11
ffff
hxfxfxfxf
fxfxfxfx
.).f()x f( .xh
iiii
ii
ii
i-i-
January 22,2015 Thursday Repl class
Dr Azizan 10
Solution
288)125.0(
)82617.68)102(26738.139)125.0(
)875.1()2(2)125.2()2(
)()(2)()(
125.0Using
2
2
211
ffff
hxfxfxfxf
h
iiii
January 22,2015 Thursday Repl class
Dr Azizan 11
Conclusion• Both results are exact because the errors are a
function of 4th and higher derivatives which are zero for a 3rd order polynomial function.
January 22,2015 Thursday Repl class
Dr Azizan 12
Homework
ions.approximatdifferencefinitecenteredandbackwardforward,theusing
functiontheofsderivativesecondandfirsttheFind.25.0with]2,2[intervalthe
on42)(functiontheConsider 3
hxxxf