Numerical methods - lecture 7
Jiří Zelinka
Autumn 2017
Jiří Zelinka Numerical methods - lecture 7 1/16
Interpolation
x0,..., xn - given points, x,- ^ Xj for / ^ j fo,...,fn- given function values (measurments), f\ = f (x,-) 0(x) = ao0o(x) + * * * + a„0n(x) - given function depending on the parameters a0,..., a„.
Examples:
0(x) = a0 + aix + • <D(x) = a0 + aie/x +
+ 3nxn: a polynomial,
• + anemx: a trigonometric polynomial
Problem of interpolation:
find the parameters ao,..., 3n to fulfill conditions
0(x/) = /y, for / = 0,1,..., n.
Jiří Zelinka
Numerical methods - lecture 7
2/16
Interpolation by polynomials
Theorem
For given points (x,-, /y), / = 0,..., /?, x,- ^ x/ for / ^ j there exists the unique polynomial P of degree at most n with
P(x/) = /y,    / = 0,..., n.
Uniqueness:
If P^x,) = P2(x;) = fn i = 0,..., n., then Q = Pi - P2 is a polynomial of degree at most n and (?(x,) = 0, / = 0,..., /?., i.e., (? has n + 1 roots so Q must be zero polynomial.
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Existence Construction of P: We construct the polynomials /_,-:
• Li is a polynomial of degree n,
pro / ^ j pro / =7.
Points xjj 7^ / are roots of /.,-:
—1 — — |-  j - - -
/./(x) = A(x - x0)... (x - x,_i)(x - X/+i) ... (x - x„).
or
/./(x) = A;7T;(x), where 7T,-(x) = TT(X — xy)
/./(x,) = 1   =}►   A =
1
7T/(X/)'
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Jiří Zelinka Numerical methods - lecture 7
4/16
L, (x) =
7T/(X/)
n (x/
-xj) ^9)
/./ - Lagrange base polynomials Lagrange interpolation polynomial
n
n
n(x-xj)
P(x) = J2 HM = E f;
;'=0
^ n(x/-xy)
;=o
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Jiří Zelinka Numerical methods - lecture 7
5/16
Example:
Xi	-1	0	1	3
fi	-3	1	-1	1
Lo(x Li(x
L2(x
L3(x P{x
(x - 0)(x - l)(x - 3)
(-1 -	o)(-	1 -		-1	-3)
(x + i;	)(x —	1)1	[x —	3)	1
(o +1;	)(0-	1)1	[0-	3)	3
(x + i;	)(x —	0)1	[x —	3)	
(i +1;	)(1 -	0)1	[1 -	3)	
(x + i;	)(x —	0)1	[x —	1)	1
(3 +1;	)(3-	0)1	[3-	1)	
1   3      1 2
= -g*3 + -2x2 -
1
3
3
8X
1
x3-
1
2 1
24
3
4
x
=  -3L0{x) + Uix) - L2{x) + L3{x) = x3 - 3x2 + 1
Jiří Zelinka
Numerical methods - lecture 7
6/16
Lagrange base polynomials
Base polynomial L0
Base polynomial L
Base polynomial L2
Base polynomial L3
Jiří Zelinka
-1 -0.5 0 0.5 1 1.5
□ S" Numerical methods - lecture 7
2.5 3 3.5
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Interpolation polynomial
Jin Zelinka Numerical methods - lecture 7 8/16
Effective calculation of /.,
Calculation of one base polynomial L; is 0(n2), i.e. direct calculation of the interpolation polynomial is 0(n3).
Effective calculation:
n
u{x) = n (* - xj)
j=0
7T,(x) = uj(x) : (x — Pi(xi)
p
0(n2)
x,-) Horner scheme, O(n) Horner scheme, O(n) 0(n2)
Jiří Zelinka
□ g Numerical methods - lecture 7
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Example:
x j I -1   O   1 3
u>(x
7T0(x 7Ti(x' 7T2{x] 7T3{x]
(x + l)(x - 0)(x - l)(x - 3) = x4 - 3x3 - x2 + 3x u>(x) : (x + 1) uj(x) : (x + O) o;(x) : (x — 1) co(x) : (x — 3)
Jiří Zelinka
Numerical methods - lecture 7
10/16
Horner scheme for division uj(x) : (x — x0), i.e. lo(x) : (x + 1)
-1
1    -3-13 0
1 -4
0 0
tto(x) = x3 - 4x2 + 3x
Horner scheme for 7To(xo) = 7Tq(—1):
7t(x)	1   -4 3	0	
-1	1   -5 8	-8	
to(-1)	= -8		
Lo(x) --	_    t0(x)    _ l/x3 tto(xo)         8 V		- 4x2 + 3x)
Similarly Z_1? /_2
• • •
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Disadvantage of the Lagrange interpolation polynomial: adding a point (xn+i, Z^+i) will cause recalculation of all base polynomials /_;.
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Newton interpolation polynomial
Base functions:
4>oM = 1.
<t>i(x) = (x-xo),
<D2(x) = (x - x0)(x - xi),
<Dn(x) = (x - x0) • • • (x - x„_i). Interpolation polynomial:
Pn{x) = 3o0o(x) + • • • + an0n(x)
Adding a point (xn+1, fn+1):
Pn+i(x) = P„(x) + an+i0n+i(x)
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Calculation of parameters a,:
a i = ŕ[x0, xi,..., x,-] - divided difference
f[xil = f;
L ]•)'''•)   j+k\ Xj+k-Xj
i.e.
f[x x.l  _ ^[xi,...,x/]-f[x|D,...,x/_i]
L U? • • • ?   /J x,-—xo
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Jiří Zelinka Numerical methods - lecture 7
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P(x)   =   /r[x0] + f[xo,xi](x-xo) + /r[xo,Xi,x2](x-xo)(x-xi) + + ----h f[xo,..., x„](x - x0) • • • (x - X„_i)
Table of divided diffrerences
X;    f, f[Xi,Xi+1] f[x/,X;+i,X/+2]
xi f > > f[xb,xux2] ^
x2f2>f^ . " f[x0.....x,7]
:      :     >   f[xn-l,X„] ^ nXn-^Xn-!,*]
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Example:				
	f,			
-1	-3			
0	1	>	1+3 _ 0+1	4
1	-1	>	-1-1 1-0	
3	1	>	1+1 _ 3-1	1
f[xh X;+i, x/+2]        f[xo, Xi, x2, x3]
-2-4 1+1
1+2 _
-3
3-0
= 1
1+3 = 3+1
1
P(x)  =   -3 + 4(x + l)-3(x+l)x + l(x+l)x(x-l) = = x3-3x2 + l
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