LECTURE 3 LAGRANGE INTERPOLATION
LECTURE 3 LAGRANGE INTERPOLATION ? Fit N + 1 points with an Nth degree polynomial
CE30125 - Lecture 3
f2
f0 f1
f3
f4
g(x) f(x)
fN
x0 x1 x2 x3 x4 ... xN
? fx = exact function of which only N + 1 discrete values are known and used to establish an interpolating or approximating function gx
? gx = approximating or interpolating function. This function will pass through all specified N + 1 interpolation points (also referred to as data points or nodes).
p. 3.1
CE30125 - Lecture 3
? The interpolation points or nodes are given as:
xo fxo fo x1 fx1 f1 x2 fx2 f2
:
xN fxN fN
? There exists only one Nth degree polynomial that passes through a given set of N + 1 points. It's form is (expressed as a power series):
gx = ao + a1x + a2x2 + a3x3 + + aNxN
where ai = unknown coefficients, i = 0 N (N + 1 coefficients).
? No matter how we derive the Nth degree polynomial, ? Fitting power series ? Lagrange interpolating functions ? Newton forward or backward interpolation
The resulting polynomial will always be the same!
p. 3.2
Power Series Fitting to Define Lagrange Interpolation
CE30125 - Lecture 3
? gx must match fx at the selected data points
gxo = fo
gx1 = f1
:
gxN = fN
ao + a1xo + a2xo2 + + aNxoN = fo ao + a1x1 + a2x12 + + aNx1N = f1
:
ao + a1xN + a2xN2 + + aNxNN = fN
? Solve set of simultaneous equations
1 xo xo2 xoN ao
fo
1 x1 x12 x1N a1 = f1
:
:
1 xN xN2 xNN aN
fN
? It is relatively computationally costly to solve the coefficients of the interpolating function gx (i.e. you need to program a solution to these equations).
p. 3.3
CE30125 - Lecture 3
Lagrange Interpolation Using Basis Functions
? We note that in general gxi = fi
? Let
N
gx = fi Vix
i=0
where Vix = polynomial of degree N associated with each node i such that
Vi
xj
0 1
ij i = j
? For example if we have 5 interpolation points (or nodes)
gx3 = foVox3 + f1V1x3 + f2V2x3 + f3V3x3 + f4V4x3
Using the definition for Vixj : V0x3 = 0 ; V1x3 = 0 ; V2x3 = 0 ; V3x3 = 1 ; V4x3 = 0 ,we have:
gx3 = f3
p. 3.4
? How do we construct Vix ? ? Degree N ? Roots at xo x1 x2 xi ? 1 xi + 1 xN (at all nodes except xi ) ? Vixi = 1
CE30125 - Lecture 3
? Let Wix = x ? xox ? x1x ? x2x ? xi ? 1x ? xi + 1x ? xN ? The function Wi is such that we do have the required roots, i.e. it equals zero at nodes xo x1 x2 ... , xN except at node xi ? Degree of Wix is N ? However Wix in the form presented will not equal to unity at xi
? We normalize Wix and define the Lagrange basis functions Vix
Vix = ---x---i-x--?---?-x---x-o---o----x---xi---?--?---x-x--1-1------x--x-i---??-----xx---22--------------xx---i-?--?---x-x--i--i?--?--1-1------x--x---?i---?-x---x-i--+i---+-1---1----------x---x--?--i--?x----N-x---N----
p. 3.5
CE30125 - Lecture 3
? Now we have Vix such that Vixi equals:
Vixi = ---x---i-x--?-i---?x---o-x---o---x---i--x-?-i---?-x--1--x---1---x---i-x--?-i---?x---2-x---2----------x---i-x-?--i---?x---i-x-?--i--1?----1---1----x---i-x--?-i---?x---i-x--+-i--1-+---1----------x---i-x--?-i---?x---N-x---N----
Vixi = 1
? We also satisfy Vixj = 0 for i j e.g. V1x2 = ---x--x-2--1--?--?--x--x-o--o-----1---1------x--x-2--1-?--?---x--x-2--2-------x--x-1--2--?--?--x--x-3--3------------x--x-1--2--?--?--x--x-N--N---- = 0
? The general form of the interpolating function gx with the specified form of Vix is:
N
gx = fiVix
i=0
? The sum of polynomials of degree N is also polynomial of degree N ? gx is equivalent to fitting the power series and computing coefficients ao aN .
p. 3.6
CE30125 - Lecture 3
Lagrange Linear Interpolation Using Basis Functions
? Linear Lagrange N = 1 is the simplest form of Lagrange Interpolation
1
gx = fiVix
i=0
gx = foVox + f1V1x
where
Vox = ---x-x--o--?--?---x-x--1-1--- = ---x-x--1-1---?-?---x-x--o---
and
V1x = ---x-x--1--?--?---x-x--o-o---
1.0 V0 (x) x0
V1(x)
(x) x1
p. 3.7
Example
? Given the following data:
xo = 2
fo = 1.5
x1 = 5
f1 = 4.0
Find the linear interpolating function gx
? Lagrange basis functions are:
Vox
=
-5----?-----x3
and
V1x
=
-x----?----2-3
? Interpolating function g(x) is:
gx = 1.5Vox + 4.0V1x
CE30125 - Lecture 3
p. 3.8
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