View
30
Download
0
Category
Preview:
Citation preview
CE 341/441 - Lecture 6 - Fall 2004
known and used to estab-
tion will pass through allints or nodes).
p. 6.1
LECTURE 6
LAGRANGE INTERPOLA TION
• Fit points with an degree polynomial
• = exact function of which only discrete values arelish an interpolating or approximating function
• = approximating or interpolating function. This funcspecified interpolation points (also referred to asdata po
N 1+ Nth
f1
x0
g(x)f(x)
f0
f2
f3 f4fN
x1 x2 x3 x4 xN...
f x( ) N 1+g x( )
g x( )N 1+
CE 341/441 - Lecture 6 - Fall 2004
through a given set of
).
N 1+
p. 6.2
• The interpolation points or nodes are given as:
:
• There exists only one degree polynomial that passespoints. It’s form is (expressed as a power series):
where = unknown coefficients, ( coefficients
• No matter how we derive the degree polynomial,
• Fitting power series
• Lagrange interpolating functions
• Newton forward or backward interpolation
The resulting polynomial will always be the same!
xo f xo( ) f o≡
x1 f x1( ) f 1≡
x2 f x2( ) f 2≡
xN f xN( ) f N≡
Nth
g x( ) ao a1x a2x2 a3x3 … aN xN+ + + + +=
ai i 0 N,= N 1+
Nth
CE 341/441 - Lecture 6 - Fall 2004
s of the interpolating func-ations).
f o
f 1=
f N=
p. 6.3
Power Series Fitting to Define Lagrange Interpolation
• must match at the selected data points⇒
⇒
⇒
: :
⇒
• Solve set of simultaneous equations
• It is relatively computationally costly to solve the coefficienttion (i.e. you need to program a solution to these equ
g x( ) f x( )
g xo( ) f o= ao a1xo a2xo2 … aNxo
N+ + + + =
g x1( ) f 1= ao a1x1 a2x12
+ + … aNx1N
+ +
g xN( ) f N= ao a1xN+ a2xN2 … aNxN
N+ + +
1 xo xo2 … xo
N
1 x1 x12 … x1
N
… … … … …
1 xN xN2 … xN
N
ao
a1
:
aN
f o
f 1
:
f N
=
g x( )
CE 341/441 - Lecture 6 - Fall 2004
h node such that
; ;
i
V4 x3( )
x3( ) 0= V3 x3( ) 1=
p. 6.4
Lagrange Interpolation Using Basis Functions
• We note that in general
• Let
where = polynomial of degree associated with eac
• For example if we have 5 interpolation points (or nodes)
Using the definition for : ; ;
,we have:
g xi( ) f i=
g x( ) f i Vi x( )i 0=
N
∑=
Vi x( ) N
Vi x j( )0 i j≠1 i = j
≡
g x3( ) f oVo x3( ) f 1V1 x3( ) f 2V2 x3( ) f 3V3 x3( ) f 4+ + + +=
Vi xj( ) V0 x3( ) 0= V1 x3( ) 0= V2
V4 x3( ) 0=
g x3( ) f 3=
CE 341/441 - Lecture 6 - Fall 2004
)
ots, i.e. it equals zero at
nity at
ns
)
xi
x( )
x xN–( ))… xi xN–( )------------------------------
p. 6.5
• How do we construct ?
• Degree
• Roots at (at all nodes except
•
• Let
• The function is such that wedo have the required ronodes except at node
• Degree of is
• However in the form presented will not equal to u
• We normalize and define the Lagrange basis functio
Vi x( )
N
xo x1 x2 …xi 1– xi 1+ … xN, , , , , , xi
Vi xi( ) 1=
Wi x( ) x xo–( ) x x1–( ) x x2–( )… x xi 1––( ) x xi 1+–( )… x xN–(=
Wi
xo x1 x2 ... , xN, , , xi
Wi x( ) N
Wi x( )
Wi x( ) Vi
Vi x( )x xo–( ) x x1–( ) x x2–( )… x xi 1––( ) x xi 1+–( )…
xi xo–( ) xi x1–( ) xi x2–( )… xi xi 1––( ) xi xi 1+–(-----------------------------------------------------------------------------------------------------------------------------=
CE 341/441 - Lecture 6 - Fall 2004
⇒
specified form of is:
of degree
ting coefficients .
… xi xN–( )xi xN–( )
----------------------------
N)
N)------- 0=
Vi x( )
N
ao … aN, ,
p. 6.6
• Now we have such that equals:
• We also satisfy
e.g.
• The general form of the interpolating function with the
• The sum of polynomials of degree is also polynomial
• is equivalent to fitting the power series and compu
Vi x( ) Vi xi( )
Vi xi( )xi xo–( ) xi x1–( ) xi x2–( )… xi xi 1––( ) 1( ) xi xi 1+–( )
xi xo–( ) xi x1–( ) xi x2–( )… xi xi 1––( ) xi xi 1+–( )…----------------------------------------------------------------------------------------------------------------------------------=
Vi xi( ) 1=
Vi xj( ) 0 for i j≠=
V1 x2( )x2 xo–( ) 1( ) x2 x2–( ) x2 x3–( )⋅ … x2 x–(x1 xo–( ) 1( ) x1 x2–( ) x1 x3–( )… x1 x–(
---------------------------------------------------------------------------------------------------=
g x( )
g x( ) f iVi x( )i 0=
N
∑=
N
g x( )
CE 341/441 - Lecture 6 - Fall 2004
nterpolation
x xo–( )x1 xo– )-------------------
p. 6.7
Lagrange Linear Interpolation Using Basis Functions
• Linear Lagrange is the simplest form of Lagrange I
⇒
where
and
N 1=( )
g x( ) f iVi x( )i 0=
1
∑=
g x( ) f oVo x( ) f 1V1 x( )+=
Vo x( )x x1–( )xo x1–( )
---------------------x1 x–( )x1 xo–( )
---------------------= = V1 x( )(--=
x0
(x)
V0 (x)
x1
V1(x)1.0
CE 341/441 - Lecture 6 - Fall 2004
p. 6.8
Example
• Given the following data:
Find the linear interpolating function
• Lagrange basis functions are:
and
• Interpolating functiong(x) is:
xo 2= f o 1.5=
x1 5= f 1 4.0=
g x( )
Vo x( ) 5 x–3
-----------= V1 x( ) x 2–3
-----------=
g x( ) 1.5Vo x( ) 4.0V1 x( )+=
CE 341/441 - Lecture 6 - Fall 2004
(x)
p. 6.9
x0 = 2
1.5 V0 (x)4
2
4
2
x1 = 5x
4.0 V1(x)
x0 = 2 x1 = 5x
x0 = 2 x1 = 5
g(x) = 1.5 V0(x) + 4.0V1
CE 341/441 - Lecture 6 - Fall 2004
p. 6.10
Lagrange Quadratic Interpolation Using Basis Functions
• For quadratic Lagrange interpolation,N=2
⇒
where
g x( ) f i Vi x( )i 0=
2
∑=
g x( ) f oVo x( ) f 1V1 x( ) f 2V2 x( )+ +=
Vo x( )x x1–( ) x x2–( )
xo x1–( ) xo x2–( )-------------------------------------------=
V1 x( )x xo–( ) x x2–( )
x1 xo–( ) x1 x2–( )-------------------------------------------=
V2 x( )x xo–( ) x x1–( )
x2 xo–( ) x2 x1–( )-------------------------------------------=
CE 341/441 - Lecture 6 - Fall 2004
are defined such that the
. Thus:i
0
1
2
p. 6.11
• Note that the location of the roots of , and
basic premise of interpolation is satisfied, namely that
x0
V0 (x)
1.0
x1
xx2
V1(x) V2(x)
V0 x( ) V1 x( ) V2 x( )g xi( ) f=
g xo( ) Vo xo( ) f o V1 xo( ) f 1 V2 xo( ) f 2+ + f= =
g x1( ) Vo x1( ) f o V1 x1( ) f 1 V2 x1( ) f 2+ + f= =
g x2( ) Vo x2( ) f o V1 x2( ) f 1 V2 x2( ) f 2+ + f= =
CE 341/441 - Lecture 6 - Fall 2004
p. 6.12
Example
• Given the following data:
Find the quadratic interpolating function
• Lagrange basis functions are
• Interpolating functiong(x) is:
xo 3= f o 1=
x1 4= f 1 2=
x2 5= f 2 4=
g x( )
Vo x( ) x 4–( ) x 5–( )3 4–( ) 3 5–( )
----------------------------------=
V1 x( ) x 3–( ) x 5–( )4 3–( ) 4 5–( )
----------------------------------=
V2 x( ) x 3–( ) x 4–( )5 3–( ) 5 4–( )
----------------------------------=
g x( ) 1.0Vo x( ) 2.0V1 x( ) 4.0V2 x( )+ +=
CE 341/441 - Lecture 6 - Fall 2004
2(x)
p. 6.13
1.0 V0 (x)1.0x
4.0 V2(x)
x1 = 4x0 = 3 x2 = 5
2.0
4.0
x1 = 4x0 = 3 x2 = 5
x1 = 4x0 = 3 x2 = 5
x
x
2.0 V1(x)
x1 = 4x0 = 3 x2 = 5
4.0g(x) = 1.0 V0(x) + 2.0V1(x) + 4.0V
CE 341/441 - Lecture 6 - Fall 2004
with )f x( ) xln=
) x x3–( )
2) x1 x3–( )--------------------------
x1) x x2–( )x1) x3 x2–( )
---------------------------------
p. 6.14
Lagrange Cubic Interpolation Using Basis Functions
• For Cubic Lagrange interpolation,N=3
Example
• Consider the following table of functional values (generated
• Find as:
0 0.40 -0.916291
1 0.50 -0.693147
2 0.70 -0.356675
3 0.80 -0.223144
i xi f i
g 0.60( )
g x( ) f o
x x1–( ) x x2–( ) x x3–( )xo x1–( ) xo x2–( ) xo x3–( )
----------------------------------------------------------------- f 1
x xo–( ) x x2–(x1 xo–( ) x1 x–(
---------------------------------------+=
f 2
x xo–( ) x x1–( ) x x3–( )x2 xo–( ) x2 x1–( ) x2 x3–( )
----------------------------------------------------------------- f 3
x xo–( ) x –(x3 xo–( ) x3 –(
--------------------------------+ +
CE 341/441 - Lecture 6 - Fall 2004
0.80– )0.80– )
------------------
0)0)----
0)0)----
0)0)----
p. 6.15
⇒
g 0.60( ) 0.9162910.60 0.50–( ) 0.60 0.70–( ) 0.60(0.40 0.50–( ) 0.40 0.70–( ) 0.40(
------------------------------------------------------------------------------⋅–=
0.6931470.60 0.40–( ) 0.60 0.70–( ) 0.60 0.8–(0.50 0.40–( ) 0.50 0.70–( ) 0.50 0.8–(
--------------------------------------------------------------------------------------------⋅–
0.3566750.60 0.40–( ) 0.60 0.50–( ) 0.60 0.8–(0.70 0.40–( ) 0.70 0.50–( ) 0.70 0.8–(
--------------------------------------------------------------------------------------------⋅–
0.2231440.60 0.40–( ) 0.60 0.50–( ) 0.60 0.7–(0.80 0.40–( ) 0.80 0.50–( ) 0.80 0.7–(
--------------------------------------------------------------------------------------------⋅–
g 0.60( ) 0.509976–=
CE 341/441 - Lecture 6 - Fall 2004
given by:
omial
p. 6.16
Err ors Associated with Lagrange Interpolation
• Using Taylor series analysis, the error can be shown to be
where
derivative of w.r.t. evaluated at
• Notes
• If = polynomial of degree where , then
⇒ for all x
Therefore will be an exact representation of
e x( ) f x( ) g x( )–=
e x( ) L x( ) f N 1+( ) ξ( )= xo ξ xN≤ ≤
fN 1+ ξ( ) N 1
th+= f x ξ
L x( )x xo–( ) x x1–( )… x xN–( )
N 1+( )!----------------------------------------------------------------- an N 1
thdegree polyn+= =
f x( ) M M N≤
fN 1+( )
x( ) 0= e x( ) 0=
g x( ) f x( )
CE 341/441 - Lecture 6 - Fall 2004
is small and if
.
.) formulae
um near the edges
rge outside of the interval.
fN 1+( )
x( )
--
p. 6.17
• Since in general is not known, if the intervaldoes not change rapidly in the interval
where
• can be estimated by using Finite Difference (F.D
• will significantly effect the distribution of the error
• is a minimum at the center of and a maxim
• e.g. using 6 point interpolation looks like:
• at all data points
• largest . becomes very la
ξ xo xN,[ ]
e x( ) L x( ) f N 1+( ) xm( )≈ xm
xo xN+
2-----------------=
fN 1+( )
L x( )
L x( ) xo xN,[ ]
L x( )
0 1 2 3 4 5
L x( ) 0=
L x( ) 0 x 1≤ ≤ 4 x 5≤ ≤ L x( )
CE 341/441 - Lecture 6 - Fall 2004
s the maximum error within
↑
↓
and thus ↑
ur!!
d vary
xN≤
x x0 x xN≤ ≤
emax x0 x xN≤ ≤
p. 6.18
• As the size of the interpolating domain increases, so doethe interval
↑ ⇒ ↑ ⇒
• As increases from a small value, ↓ ⇒
• However as ⇒ ↑ for a given
• Therefore convergence as↑ does not necessarily occ
• Properties of will also influence error as an
D xN xo–= Lmax x0 x xN≤ ≤emax x0 x≤
N Lmax x0 x xN≤ ≤ema
N NCRIT> Lmax x0 x xN≤ ≤xo xN,[ ]
N
fN 1+( ) ξ( ) D N
CE 341/441 - Lecture 6 - Fall 2004
at (usually wex) x( )ln=
80)------ f3 1+( )
0.6( )
p. 6.19
Example
• Estimate the error made in the previous example knowing thdo not have this information).
⇒
⇒
⇒
f (
e x( ) L x( ) fN 1+( )
xm( )≈
e x( )x xo–( ) x x1–( ) x x2–( ) x x3–( )
3 1+( )!----------------------------------------------------------------------------- f
3 1+( )xm( )≈
e 0.60( ) 0.60 0.40–( ) 0.60 0.50–( ) 0.60 0.70–( ) 0.60 0.–(3 1+( )!
--------------------------------------------------------------------------------------------------------------------------≈
e 0.60( ) 0.000017 f4( )
0.6( )=
CE 341/441 - Lecture 6 - Fall 2004
nction itself
a Finite Difference (F.D.)
p. 6.20
• We estimate the fourth derivative off(x) using the analytical fu
⇒
⇒
⇒
⇒
⇒
• Therefore
• Exact error is computed as:
Therefore error estimate is excellent
• Typically we wouldalso have to estimate using
approximation (a discrete differentiation formula).
f x( ) xln=
f1( )
x( ) x1–
=
f2( )
x( ) x2–
–=
f3( )
x( ) 2x3–
=
f4( )
x( ) 6x4–
–=
f4( )
0.6( ) 46.29–=
e 0.60( ) 0.00079–=
E x( ) 0.60( ) g 0.60( )–ln 0.00085–= =
fN 1+( )
xm( )
CE 341/441 - Lecture 6 - Fall 2004
ints.
l→ find coefficients by
hrough data points→r.
) pointi
ch that it equals unity at thedata points and nonzero in-
N 1+
p. 6.21
SUMMARY OF LECTURES 5 AND 6
• Linear interpolation passes a straight line through 2 data po
• Power series→ data points→ degree polynomiasolving a matrix
• Lagrange Interpolation passes an degree polynomial tUse specialized nodal functions to make finding easie
where
= the interpolating function approximatingf(x)
fi = the value of the function at the data (or interpolation
= the Lagrange basis function
• Each Lagrange polynomial or basis function is set up sudata point with which it is associated, zero at all otherbetween.
N 1+ Nth
Nth
g x( )
g x( ) f iVi x( )i 0=
N
∑=
g x( )
Vi x( )
CE 341/441 - Lecture 6 - Fall 2004
p. 6.22
• For example whenN 2= 3 data points→
V0 V1 V2
0 1 2
g x( ) f oVo x( ) f 1V1 x( ) f 2V2 x( )+ +=
f0
f1
f2 g(x)
CE 341/441 - Lecture 6 - Fall 2004
with
(or at some point in
1=
m)
mial
p. 6.23
• Linear interpolation is the same as Lagrange Interpolation
• Error estimates can be derived but depend on knowingthe interval).
where
derivative of w.r.t. evaluated at
N
fN 1+( )
x(
e x( ) L x( ) f N 1+( ) ξ( )= xo ξ xN≤ ≤
fN 1+ ξ( ) N 1
th+= f x ξ
L x( )x xo–( ) x x1–( )… x xN–( )
N 1+( )!----------------------------------------------------------------- an N 1
thdegree polyno+= =
Recommended