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Philadelphia University
Engineering Faculty Mechanical Engineering Department
Eng. Laith Batarseh
Introuction To numerical methods
Introduction to numerical methods
Numerical methods are those in which the mathematical
problem is reformulated so it can be solved by arithmetic
operations.
We use numerical methods when:
Geometry is very complex
There is no way to find exact solution
Mathematical model general form:
Function
ForcingParameters
Variable
tIndependenf
Variable
Dependent,,
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Introduction to numerical methods
Problem Definition
Mathematical Model
Numeric Or Graphic Results
Implementation
Physical problem + Assumptions + Boundary or Initial
Conditions
Transform the physical problem into mathematical equation or
equations and define the solution desired to be obtained
Solve the Mathematical model to find the desired solution and
represent in by numeric or graphic means
Explain and discus the mathematical solution and refer it back
to the physical problem
Introduction to numerical methods
Significant Figures: are the digits of a number we can use
with confidence.
Significant Figures = number of certain digits + one digit
Examples:
Significant Figures
0 1 2 3 4 5
26.5 mm
Uncertain
Digits
Certain
Digits
3 Sign. Figs
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Introduction to numerical methods
Special Cases:
2.55 & 25.5 & 255 & 2550 & 25500 have 3 Sig. Figs.
2.55 & 0.255 & 0.0255 & 0.00255 have 3 Sig. Figs.
Significant Figures
Representation : 2.55 x 101 & 2 & 3 & 4
Representation : 2.55 x 10-1 & -2 & -3 & -4
Introduction to numerical methods
Accuracy refers to how closely a computed or measured value
agree with the true value.
Precision refers to how closely individual computed or measured
value agree with each other.
Bias (inaccuracy) is the opposite of accuracy. Systematic
deviation from true value
Uncertainty (imprecision) is the opposite of precision.
Magnitude of scatter.
Example:
Accuracy and Precision
Inaccurate &
imprecise
accurate &
imprecise
Inaccurate &
precise
accurate &
precise
Numerical
methods must be
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Introduction to numerical methods
Error represents the deviation from the truth or true value
There are two types of error may appear when numerical
methods are used: Round – off error andTruncation error.
Both errors can be represented mathematically by:
Errors Definitions
%100xTrueValue
ionApproximatTrueValue
ionApproximatTrueValueE
t
t
%100
Pr
Pr
xionApproximatCurrent
ionApproximateviousionApproximatCurrent
ionApproximateviousionApproximatCurrentE
t
a
Et = True error
εt = relative true error
Ea = Approximate error
εa= relative Approximate error
Introduction to numerical methods
Round – off error: Results form chopping off some
significant figures. This type of error is found in computer
processes.
Example:
π = 3.14159265….
Computer limitations in representation of numeral
quantity create such error.
Computer reduce such problem by rearranging number
presentation : 25560000 → 2.556 x107
Round – off error
Round – off error
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Introduction to numerical methods
Truncation error: Results form using approximationmethods instead of exact mathematical procedures to findthe solution
Example:
Assume f(x) = x2+x+1 , then f(5) = 52 +5+1 = 31
Assume you used the 1st order Taylor series for
example to find the value of f(5) and the results was:
f(5) =f(4)+f’(4)(5-4) = 30 assume xi = 4 , xi+1 = 5
Error = 31 – 30 = 1 (this is a truncation error )
Truncation error
Truncation Error
Taylor series formula:
: the reminder ofTaylor series
Where:
h = xi+1 – xi (step size)
xi : initial independent value
xi+1:next independent value
ζ: is a value lies somewhere between xi+1 and xi
n: Taylor series approximation order
Taylor series
n
n
i
n
iiiii R
n
hxfhxfhxfhxfxfxf
!
)(...
!3
)('''
!2
)('')(')()(
)(32
1
! 1
)(1)1(
n
hfR
nn
n
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Truncation Error
Example#1:
If and xi = 0 xi+1 = 1
Then theTaylor series expansion for
Note: when applying Taylor series to estimate polynomials, the number
of terms that gives the exact solution equals the order of approximation
and the rest term will be equal zero (ie. f(n+1) =0)
Taylor series
52635)(234 xxxxxf
!4
)(
!3
)('''
!2
)('')(')()(
4432
1
hxfhxfhxfhxfxfxf iii
iii
Truncation Error
Example#1(cont):
Taylor series
1 & 0 52635)( 1
234 ii xxxxxxxf
!4
)(
!3
)('''
!2
)('')(')()(
4432
1
hxfhxfhxfhxfxfxf iii
iii
5)0( ixf
120)0( 120)(
18)0( 18120)(
12)0( 121860)(
2)0( 212920)(
)4()4(
)3()3(
)2(2)2(
)1(23)1(
fxf
fxxf
fxxxf
fxxxxf
21
!4
0
!3
0
!2
0)0()0()1(:order 4
16!3
0
!2
0)0()0()1(:.order 3
13!2
0)0()0()1(:order 2
7)0()0()1(:order 1
5)0()1(:order 0
)4()4()3()3()2()2()1(th
)3()3()2()2()1(rd
)2()2()1(nd
)1(st
th
hfhfhfhfff
hfhfhff
hfhfff
hfff
ff
21)1( 1 ixf
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Truncation Error
Example#1(cont):
Taylor series
%0%10021
2121:order 4
%81.23%10021
1621:.order 3
%09.38%10021
1321:order 2
%67.66%10021
721:order 1
%19.76%10021
521:order 0
th
rd
nd
st
th
x
x
x
x
x
t
t
t
t
t
21)1( 1 ixf
Note : as the order of Taylor approximation increase, the
accuracy increase too
Truncation Error
Example#2:
and xi = π/4 xi+1 =π/3
Solution:
Taylor series
)cos()( xxf
Order (n) f(n)(x) f (π/3) εt
0 Cos(x) 0.707106781 41.4
1 -sin(x) 0.521986659 4.4
2 -cos(x) 0.497754491 0.449
3 Sin(x) 0.499869147 2.62x10-2
4 Cos(x) 0.500007551 1.51x10-3
5 -sin(x) 0.500000304 6.08x10-5
6 -cos(x) 0.499999988 2.44x10-6
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Truncation Error
Taylor series reminder:
Assume 0th orderT.S:
It is inconvenient to deal with the remainder as infinite series . So, let us start
with the 1st term of the remainder :
derivative mean – value theorem states that if a function f(x) and its derivative
are continues over an interval from xi and xi+1, then there exists at least one point
on the function has a slope, designated by f’(ξ), that is parallel to the line joiningf(xi) and f(xi+1).
Taylor series reminder
! 1
)(
!
)(...
!2
)('')(')()(
1)1()(2
1
n
hf
n
hxfhxfhxfxfxf
nnn
i
n
iiii
...!3
)('''
!2
)('')(' ),()(
32
1
hxfhxfhxfRxfxf ii
inii
Rn
hxfR in )('
Truncation Error
Graphical Presentation
Taylor series reminder
Rnh
RSlope o
xi Xi+1
F(x)
x
2
1!2
''
'
'
hf
R
hfR
h
Rf
o
o
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Truncation Error
Example:
Let us take the 1st order approximation inTaylor series:
Rearrange the terms :
Note that the 1st part is used to approximate the 1st derivative in
numerical methods
Using Taylor series to estimate truncation error
111 ' Rxxxfxfxf iiiii
iiii
iii
xx
R
xx
xfxfxf
1
1
1
1 '
First-order
Approximation
Truncation
Error
Truncation Error
Example(cont):
The reminder represents the truncation error and equal to
Or:
O(xi+1-xi): error order notification and it means that the error is in the
order of h so, if h is halve, then the error is halve too. In similar way, if
the error was in order O(h2), its mean if the step h is halve the error
will be reduced to 1/4th of its previous value.
Using Taylor series to estimate truncation error
ii
ii
xxf
xx
R
1
1
1
!2
''
ii
ii
xxOxx
R
1
1
1
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Numerical Differentiation
First derivative
Formula derivation
Assume we take the 1st orderTaylor series terms
Rearrange to separate the first derivative:
Δfi : 1st forward difference.
Δf/h: 1st finite forward divided difference.
Forward Difference
111 ' Rxxxfxfxf iiiii
hOh
fxxO
xx
xfxfxf i
ii
ii
iii
1
1
1'
Numerical Differentiation
First derivative
Formula derivation
Expand 2nd orderTaylor series terms backward to get:
Truncate above equation after the 1st term and rearrange to separatethe first derivative:
: 1st Backward difference.
: 1st finite backward divided difference.
Backward Difference
...!2
'' ' 2
1 hxf
hxfxfxf iiii
hO
h
fxxO
h
xfxfxf ii
iii
11
1'
hf /1
1f
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Numerical Differentiation
First derivative
Formula derivation
Centered Difference
...!2
'' ' 2
1 hxf
hxfxfxf iiii
...
!2
'' ' 2
1 hxf
hxfxfxf iiii
211
3
11
2...
62' hO
h
xfxfxf
h
xfxfxf iiiii
i
Numerical Differentiation
Graphical representation
Xi-1 Xi+1
F(x)
xxi
True solution
Backward
Forward
Centered
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Examples
Example #1
Given information
assume you have the following functions :
Find the value of f(2) using Taylor series 3rd order expansion. Let
h=0.5 and find the relative true error (εt)
Taylor series
1
1)( .
1)( .
1
1)( .
3
34
34
2
34
x
xxxfc
xxxfb
x
xxxfa
Examples
Example #1
Solution
Taylor series 3rd order expansion (n=3) is obtained as:
The exact solution is found as:
Taylor series
312
122)2( .
3122)2( .
312
122)2( .
3
34
34
2
34
fc
fb
fa
!3
)(
!2
)()()()(
3)3(2)2()1(
1
hxfhxfhxfxfxf ii
iii
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Examples
Example #1
Solution
Taylor series
function f(1)(1.5) f(2)(1.5) f(3)(1.5) f (2)
a 0.2400 9.8080 0.2400 2.5008
b 2.0587 2.9046 2.0587 2.9944
c 2.2359 4.5990 2.2359 3.0027
0027.3!3
5.0*)5.1(
!2
5.0*)5.1(5.0*)5.1()5.1()2( .
9944.2!3
5.0*)5.1(
!2
5.0*)5.1(5.0*)5.1()5.1()2( .
5008.2!3
5.0*)5.1(
!2
5.0*)5.1(5.0*)5.1()5.1()2( .
3'''
3
2''
3'
333
3'''
2
2''
2'
222
3'''
1
2''
1'
111
fffffc
fffffb
fffffa
Examples
Example #2
Given information
assume you have the following functions :
Find the value of f’(2) using forward, backward and centered finite
divided difference
Assume xi-1 = 1.5 xi = 2 xi+1 = 2.5
Finite divided difference
1
1)( .
1)( .
1
1)( .
3
34
34
2
34
x
xxxfc
xxxfb
x
xxxfa
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Examples
Example #2
Solution
forward finite divided difference:
Assume: xi = 2 xi+1 = 2.5
Finite divided difference
5.9869
25.2
25.2)2(' .
3.8869 25.2
25.2)2(' .
3.309525.2
25.2)2(' .
fffc
fffb
fffa
ii
iii
xx
xfxfxf
1
1'
Examples
Example #2
Solution
backward finite divided difference:
Assume: xi-1 = 1.5 xi = 2
Finite divided difference
3.4303
25.2
5.12)2(' .
2.7213 25.2
5.12)2(' .
1.7000 25.2
5.12)2(' .
fffc
fffb
fffa
h
xfxfxf ii
i1'
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Examples
Example #2
Solution
centered finite divided difference:
Assume: xi-1 = 1.5 xi+1 = 2.5
Finite divided difference
4.708625.22
5.15.2)2(' .
3.304125.22
5.15.2)2(' .
2.5048 25.22
5.15.2)2(' .
fffc
fffb
fffa
h
xfxfxf ii
i2
' 11