Week01_Introduction to Numerical
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Transcript of Week01_Introduction to Numerical
-
9/11/2012
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WEEK 1
Introduction to Numerical Methods Mathematical modeling
Approximation and round off errors
Truncation errors and Taylor Series 2
At the end of this topic, the students will be able:
To describe numerical techniques as
compared to analytical methods
To use Taylor series expansion to approximate
a function
To perform error analysis associated with
numerical methods
LESSON OUTCOMES
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4
Why Numerical Method?
Could handle large systems of equations, nonlinearity and complex geometries that is not
common
It provide approximate solutions to many of the engineering problems.
Powerful analysis tool in problem solving and understanding problem in mathematical language
Techniques by which mathematical problems are formulated, so that they can be solved with
arithmetic operations
The role of numerical method in solving engineering problem:
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What is Numerical Method
PROBLEM
FORMULATION
Fundamental laws are used to
develop mathematical
equations that can represent
the specific problem
SOLUTION
Suitable numerical methods
are then selected to solve
the mathematical equations
INTERPRETATION
The results obtained can
then be used to
predict/analyze/understand
the specific problem better
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Is the use of mathematics to
Describe real world phenomena
Investigate important questions about the observed world
Explain real world phenomena
Test ideas
Make predictions about the real world
The real world refers to Engineering Physics
Physiology Ecology
Wildlife management Chemistry
Economics Sports
Etc
Mathematical modeling
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A mathematical model is represented as a functional relationship of the form
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Dependent variable
Observed behaviour/state/phenomenon of a system
Characteristic that reflects behaviour or state of the system i.e. y, f(x), f(t)
Independent variable
Dimension that determine a system i.e. time, t , x
Parameter
Quantity that serves to relate to functions and variables Reflective of the systems properties or composition
Forcing functions
External influence that acts on system i.e. acceleration gravity, g
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Example
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Apply Newtons second law, F = ma
And also can write as
Fdt
xdm
or
Fdt
dvm
2
2
Eq. relates a linear position x (dependent variable) to the applied
force, F (forcing function) and the time, t (independent variable). The mass, m is the only parameter in the above model.
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Example of mathematical modeling
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Example
Assume that interested to predict the velocity of the falling parachutist with time
Use fundamental knowledge to find a mathematical equation correlates the velocity
to the various forces acting on the parachutist
Newtons 2nd law of Motion
the time rate change of momentum of a body is equal to the resulting force acting on
it.
The model is formulated as
F = ma
F=net force acting on the body (N)
m=mass of the object (kg)
a=its acceleration (m/s2)
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m
F
dt
dv
dt
dv
aon accelerati
resistanceair of force upward
gravity of force downward
UF
DF
UFDFF
cvUF
mgDF
vm
cg
m
cvmg
dt
dv
Model relates acceleration of falling
object to the forces acting on it,
(differential equation)
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Exact or analytical solution: it exactly satisfies the original equation
dependent variable
independent variable
forcing functions
parameter
t,s v,(m/s)
0 0.00
2 16.40
4 27.77
6 35.64
8 41.10
10 44.87
12 47.87
53.39
Analytical solution of the parachutist
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Unfortunately, there are many mathematical
models that cannot be solved exactly.
Numerical solution that approximates the exact solution
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)(1
)()1
(
1
)()1
(
it
ii
it
it
ii
it
it
vm
cg
tt
vv
tt
vv
t
v
dt
dv
vm
cg
m
cvmg
dt
dv
)( 1)()()( 1 iittt ttvm
cgvv
iii
The use of finite difference to approximate the
first derivative of v with respect to t
Approximate or
numerical solution
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t,s v,(m/s)
0 0.00
2 19.60
4 32.00
6 39.85
8 44.82
10 47.97
12 49.96
53.39
Numerical solution
)( 1)()()( 1 iittt ttvm
cgvv
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Comparison between the exact and
numerical solution
Approximation and Roundoff Errors
Significant figures
98
98.09
0.0098
Numbers to be used in confidence
2 significant figures
4 significant figures
2 significant figures
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Important of significance figures in numerical methods:
Numerical methods yield approximate results, therefore, need to develop criteria to specify the confident in
approximate result.
Although quantities such as , e, or 7 represent specific quantities, they cannot be expressed exactly by a limited
number of digits. Computers retain only a finite number of
significant figures.
= 3.141592653589793238462643
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Accuracy and Precision
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Inaccurate & imprecise
accurate & precise Inaccurate & precise
accurate & imprecise
Increasing accuracy In
cre
as
ing
pre
cis
ion
How closely a computed
or measured value agree
with the true value
How closely individual
computed or measured
values agree with each
other
Error definitions
True value
Approximation
value
Error
True value = Error + Approximation value
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Truncation errors ~ result
when approximations are
used to represent exact
mathematical procedures
Round-off errors ~ result
when numbers having a
limited significant figures
are used to represent
exact numbers
t designates true percent relative error
True value = error + approximation value
True error (Et)= true value approximation
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(1)
(2)
(3)
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Calculation of errors
True value of length of a bridge is 10,000 cm.
When you measure, the length recorded is
9,999 cm. Compute the true error and true
percentage relative error of the bridge.
Answer: 1 cm and 0.01%
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However, in actual situation, true value is rarely available. Therefore, need to estimate the true value approximation
In numerical method, iterative approach is used to compute answer, in which error is estimated as the
difference between previous and current approximations.
The signs of error can be negative or positive,
Absolute value of error, |a| need to be lower than prespecified percent tolerance, s
n is significant figures. 23
(4)
(5)
Error estimates for iterative method
Suppose that we have exponential function as,
Starting with the simplest version, ex=1, add terms to estimate e0.5. Compute true (t) and approximate error (a) after each term is added until |a| falls below s , conforming to 3 significant figures. Note that true value of e0.5
is 1.648721.
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Answer
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Round-off errors
ln 2 = 0.693 147 180 559 945 309 41...
A device only shows 8
significant numbers, so
round-off error is discrepancy
introduced by omission of
significant figures.
Round-off error for this case is
0.00000000055994530941
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Results when numbers having limited significant
figures are used to represent exact numbers
Other example of roundoff error
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Truncation errors and Taylor series
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Truncation errors
Truncation error is the discrepancy introduced by the fact that numerical methods may employ approximations to represent exact
mathematical operations and quantities.
Truncation error are errors resulted from using an approximation in place of an exact mathematical procedure.
The difference between the calculated value using exact mathematical equation and approximation mathematical equation.
Zero order
First order
Second
order
nth order
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Provides a means to predict a function value at one point in terms of the
function value and its derivative at another point
Taylor series
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(6)
(7) 1
)1(
)!1(
)(
n
n
n hn
fR
Taylor series by defining a step size h = xi+1 - xi
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Let say we truncated the Taylor series expansion after zero-order term to yield
Remainder term, Rn for zero order version
Let truncate the remainder itself,
This result is still inexact because neglected second and higher order terms.
)()( 1 ii xxff
...!3!2
''' 3
)(3
2)(
)(0 hf
hf
hfR iii
xx
x
hfRix )(0
'
Remainder term, Rn, accounts for all terms from (n+1) to infinity.
It also usually expressed as:
)( 1 nn hOR
1)1(
)!1(
)(
n
n
n hn
fR
Remainder for the Taylor series Expansion
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Alternative simplification that tranforms the approximation into an equivalence based on graphical insight
derivative mean-value theorem states that if a function f(x) and its derivative are continous over interval from xi to xi+1, there exist at least one point on the function that has a slope, designated by f(), parallel to line joining f (xi) and f(xi+1)
Thus,
So,
Zero order version
First order version
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Numerical differentiation
Forward finite divided difference
Backward finite divided difference
Centered finite divided difference
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Forward finite divided difference approximation of first derivative
Backward finite divided difference approximation of first derivative
Centered finite divided difference approximation of first derivative
Where, (14)
(15)
(16)
(17)
O(h)h
f)f'(x
h
R
h
)f(x)f(x)f'(x
ii
iii
111 ii xxhWhere,
11 iiii xxxxhWhere,
36
Use forward and backward difference approximations of O(h) and a centered
difference approximation of O(h2) to estimate the first derivative of
2.125.05.015.01.0f 234 xxxx)(xi
at x = 0.5 using a step size h = 0.5. Repeat using h = 0.25. Also calculate the
true percent relative error for each approximation.
Example
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Exercise
Use forward and backward and a centered difference to estimate the first
derivative of the function
7.08.01.05.0f 23 xxx)(xi
at x = 0.5 using a step size h=0.5. Repeat using h = 0.25. Also calculate the
true percent relative error for each approximation.
Ans:
h=0.5
FDM:1.525 41.9%
BDM:0.875 18.60%
CDM:1.200 11.63%
h=0.25
FDM:1.26875 18.02%
BDM:0.94375 12.21%
CDM:1.10625 2.91%
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Second forward finite difference approximation of higher derivatives
Second backward finite difference approximation of higher derivatives
Second centred finite difference approximation of higher derivatives
Higher derivatives
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Error propagation This section is to study how errors in numbers can
propagate through mathematical functions. If we multiply
two numbers that have errors, we would like to estimate
the error in the product.
If a function f is dependent on
(a) a single independent variable x : f(x)
(b) two independent variables x and y : f(x, y)
(c) several independent variables x1, x2, x3, ... ,xn : f(x1 , x2 ,..., xn ).
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Function of a single variable.
Let x be the true value and
x* be an approximate value of x
Then, TSE for f(x) computed near f(x*) is given by
...*)(
2
*)(''*)(* 2 xx
xfxx)f'(xf(x*)f(x)
Truncating after the first derivative term and rearranging the remaining terms
to give
where
**)'
*)*)('
x(xff(x*)
xx(xff(x*)f(x)
xxxx
xfxf(xf
t variableindependen oferror theof estimatean is **
function theoferror theof estimatean is *)()(*)'
Eq.(21) provides 2 capabilities:
1. to approximate the error in f(x) knowing its derivative.
2. to approximate the error in the independent variable x.
(21)
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Example
Given a value of x* = 2.5 with an error of x* = 0.01, estimate the
resulting error in the function, f(x)=x3.
Solution
187506251552
atpredict thcan it ,625.15)5.2( Because
1875.0)01.0()5.2(3
So,
**)'
2
. .).f(
f
f(x*)
x(xff(x*)
Or the true value lies between 15.4375 and 15.8125. In fact, if x ~2.49,
f(x) could be 15.4382 and if x ~ 2.51, it would be 15.8132.
The first order error analysis provides a fairly close estimate of the true error.
42
Exercise
Knowing a value of x* = 2.0 with an error of x* = 0.01, estimate the resulting error in the function
f(x) = 0.5x3-0.1x2+0.8x-0.7
Ans: f(2.0)=4.5 0.064
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Function of a more than One variable.
(22)
(23)
Refer section 4.2.2
for examples
44
Relative error
Condition number
Refer section 4.2.3
for example
=
=
Condition no equals 1 indicates that functions relative error is identical to the relative error in
x
Condition no greater than 1 indicates relative error is amplified.
Condition no less than 1 indicates relative error is attenuated.
Function with very large values are said to be ill-conditioned.
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Total numerical errors = truncation error + round off error
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Roundoff error by increase no. of significant figures or
reduce no. of computation in analysis
Truncation error by decreasing step size (h) or increase
no. of computation in analysis
Total Numerical Error
Control numerical error
avoid subtract 2 nearly equal numbers to avoid loss of significance
Use Taylor series for truncation and roundoff error analysis
Perform numerical experiments
- repeat computation with different step size or method and compare results
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