MATH 2140 Numerical Methods
MATH 2140
Numerical Methods
Instructor:
Dr. Mohamed El-Shazly
Associate Prof. of Mechanical Design and
Tribology
melshazly@ksu.edu.sa
Office: F072
Faculty of Engineering
Mechanical Engineering Department
1
2
INTERPOLATION USING A SINGLE
POLYNOMIAL
INTERPOLATION USING A SINGLE POLYNOMIAL
•
Interpolation is a procedure in which a mathematical formula is used to
represent a given set of data points, such that the formula gives the exact
value at all the data points and an estimated value between the points. This
section shows how this is done by using a single polynomial, regardless of
the number of points.
•
As was mentioned in the previous section, for any number of points n there
is a polynomial of order n - 1 that passes through all of the points.
•
For two points the polynomial is of first order (a straight line connecting the
points).
•
For three points the polynomial is of second order (a parabola that connects
the points), and so on.
•
This is illustrated in Fig. 6-11 which shows how first, second, third, and
fourth-order polynomials connect two, three, four, and five points,
respectively
.
3
4
5
Interpolation
•
Given a sequence of
n
unique points,
(
x
i
,
y
i
)
•
Want to construct a function
f
(
x
)
that passes
through all the given points
so that …
•
We can use
f
(
x
)
to estimate the value of
y
for
any
x
inside the range
of the known base points
6
Extrapolation
•
Extrapolation
is the
process of estimating a
value of
f
(
x
)
that lies
outside the range of the
known base points.
•
Extreme care should be
exercised where one
must extrapolate.
7
Polynomial Interpolation
O
b
j
e
c
t
i
v
e
:
Given
n
+1
points, we want to find the polynomial of
order
n
that passes through all the points.
8
Polynomial Interpolation
•
The
n
th
-order polynomial that passes through
n
+1
points is
unique
, but it can be written in different
mathematical formats:
–
The conventional form
–
The Newton Form
–
The Lagrange Form
•
Useful characteristics of polynomials
–
Infinitely differentiable
–
Can be easily integrated
–
Easy to evaluate
Lagrange Interpolating Polynomials
•
Lagrange interpolating polynomials are a
particular form of polynomials that can be
written to fit a given set of data points by
using the values at the points.
•
The polynomials can be written right away
and do not require any preliminary
calculations for determining coefficients.
9
10
11
•
Equation (6.40)
•
is a linear function of x (an equation of a straight line that
connects the two points). It is easy to see that if x = x
1
is
substituted in Eq. (6.40), the value of the polynomial is
y
1
, and if x = x
2
is substituted, the value of the polynomial
is Y
i
· Substituting a value of x between the points gives
an interpolated value of y.
•
Equation (6.40) can also be rewritten in the standard
form f(x ) = a
1
x + a
0
:
12
13
14
15
16
17
18
Construct a 4
th
order polynomial in Lagrange form that
passes through the following points:
Example
We can construct the polynomial as
where
L
i
(
x
)
can be constructed separately as …
(see next page)
19
Example
Interpolation using a single polynomial in numerical methods gives insights into constructing a function passing through a set of data points. Various polynomial orders are utilized to connect data points efficiently, showcasing the importance of polynomial interpolation and extrapolation in mathematical analysis.
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Presentation Transcript
Faculty of Engineering Mechanical Engineering Department MATH 2140 Numerical Methods Instructor: Dr. Mohamed El-Shazly Associate Prof. of Mechanical Design and Tribology melshazly@ksu.edu.sa Office: F072 1
INTERPOLATION USING A SINGLE POLYNOMIAL 2
INTERPOLATION USING A SINGLE POLYNOMIAL Interpolation is a procedure in which a mathematical formula is used to represent a given set of data points, such that the formula gives the exact value at all the data points and an estimated value between the points. This section shows how this is done by using a single polynomial, regardless of the number of points. As was mentioned in the previous section, for any number of points n there is a polynomial of order n - 1 that passes through all of the points. For two points the polynomial is of first order (a straight line connecting the points). For three points the polynomial is of second order (a parabola that connects the points), and so on. This is illustrated in Fig. 6-11 which shows how first, second, third, and fourth-order polynomials connect two, three, four, and five points, respectively. 3
Interpolation Given a sequence of n unique points, (xi, yi) Want to construct a function f(x) that passes through all the given points so that We can use f(x) to estimate the value of y for any x inside the range of the known base points 5
Extrapolation Extrapolation is the process of estimating a value of f(x) that lies outside the range of the known base points. Extreme care should be exercised where one must extrapolate. 6
Polynomial Interpolation Objective: Given n+1 points, we want to find the polynomial of order n x a a x p + = 1 0 ) ( + + + 2 n a x a x 2 n n that passes through all the points. 7
Polynomial Interpolation The nth-order polynomial that passes through n+1 points is unique, but it can be written in different mathematical formats: The conventional form The Newton Form The Lagrange Form Useful characteristics of polynomials Infinitely differentiable Can be easily integrated Easy to evaluate 8
Lagrange Interpolating Polynomials Lagrange interpolating polynomials are a particular form of polynomials that can be written to fit a given set of data points by using the values at the points. The polynomials can be written right away and do not require any preliminary calculations for determining coefficients. 9
Equation (6.40) is a linear function of x (an equation of a straight line that connects the two points). It is easy to see that if x = x1is substituted in Eq. (6.40), the value of the polynomial is y1, and if x = x2is substituted, the value of the polynomial is Yi Substituting a value of x between the points gives an interpolated value of y. Equation (6.40) can also be rewritten in the standard form f(x ) = a1x + a0: 12
Example Construct a 4thorder polynomial in Lagrange form that passes through the following points: 0 0 -5 1 1 -3 2 -1 -15 3 2 39 4 -2 -9 i xi f(xi) We can construct the polynomial as = + ( ) 5 ( ) 3 ( ) 15 ( ) 39 ( ) 9 ( ) p x L x L x L x L x L x 4 0 1 2 3 4 where Li(x) can be constructed separately as (see next page) 18
0 0 -5 1 1 -3 2 -1 -15 3 2 39 4 -2 -9 i xi f(xi) Example ( )( )( + )( )( )( )( + ) ( )( )( 4 )( ) + + + + 1 1 2 2 1 1 2 2 x x x x x x x x = = ( ) L x 0 + + 0 ( 1 x 0 1 x 0 2 x 0 ) 2 ( )( + x )( 2 ) ) 2 + ( )( )( ) + + 1 2 x 2 + x 1 2 2 x x x x x = = ( ) L x 1 1 ( 0 )( x 1 1 )( 1 )( 1 6 ( )( )( x )( ) 1 ( )( )( ) + 1 2 2 1 2 2 x x x x = = ( ) L x 2 + x ( 1 x 0 x 1 1 + )( 1 x 2 )( ) 2 x 6 + ( 0 )( )( + ) ( )( )( ) + + 1 1 2 1 1 2 x x = = ( ) L x 3 + 2 ( )( x 2 x 1 )( 2 x 1 )( 2 x ) 2 24 x ( )( )( )( 2 ) ( )( )( ) + + 1 1 2 1 1 2 x x x = = ( ) L x 4 + ( 2 0 2 1 )( 1 )( 2 ) 2 24 19
