interpolation.pdf

7/30/2019 interpolation.pdf

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Interpolation MACM 316 2/13

Linear Interpolation

Equation of the line joining (1.0, 3.6788) and (2.0, 5.4134) is

p(x) = 1.9442 + 1.7346x

Use p(x) to approximate f(1.75) at an unknown point (inter-polate)

f(1.75) p(1.75) = 4.9798

Given that the exact value is f(1.75) = 5.3218, the interpola-

tion error is

|4.9798 5.3218||5.3218|

= 0.0643 (relative error)

We could extrapolate outside the interval [1, 2] to obtain

p(2.25) = 5.8471

. . . but this is usually inaccurate.

Its typically safer to derive the linear interpolant on [2, 2.5]:

q(x) = 6.5458 0.5662x = q(2.25) = 5.2719

Interpolation vs. extrapolation

1 1.5 2 2.5

4

4.5

5

5.5

x

y

October 13, 2008 c Steven Rauch and John Stockie


3/13


Higher Order Interpolation

Note: We assume the original data comes from a smooth

function, but linear interpolation on more than two points is

not smooth.

How do we get a smooth approximation for f(x)???

= use higher order polynomials!

Suppose we have a list of n + 1 points

{(x0, y0), (x1, y1), . . . , (xn, yn)}where yi = f(xi) for i = 0, 1, 2, . . . , n.

Theorem 3.3 (p. 108): Provided the points xi are all dis-

tinct, there exists a UNIQUE polynomial of degree n

Pn(x) = a0 + a1x + a2x2 + + anx

n

which satisfies Pn(xi) = yi for i = 0, 1, 2, . . . , n.Furthermore,

f(x) = Pn(x) + En(x)

where the error term

En(x) = f(n+1)

(c)(n + 1)!

(x x0)(x x1) (x xn)

for some c [x0, xn].

Pn(x) is called an interpolating polynomial for f(x).



4/13


Previous Example: The interpolating polynomial of degree 2 for

f(x) on [1, 2.5] is

P2(x) = 1.1235 + 6.3362x 1.5339x2,

since P2(1) = 3.6788, P2(2) = 5.4134 and P2(2.5) = 5.1303.

We can now approximate f(1.75) by P2(1.75) = 5.26735, which

has relative error of 0.0102 much smaller than we saw for thelinear approximation!

Quadratic vs. Linear Interpolation

1 1.5 2 2.53

3.5

4

4.5

5

5.5

x

y

f(x)P

1(x)

P2(x)

Note: Increasing the degree of the polynomial does not neces-

sarily guarantee an improved approximation.



5/13


Interpolating Polynomials The Mechanics

Suppose we have 3 points: (x0, y0), (x1, y1), (x2, y2).

We want to fit a quadratic (n = 2): y = c0 + c1x + c2x2

. Substitute:

y0 = c0 + c1x0 + c2x20

y1 = c0 + c1x1 + c2x21

y2 = c0 + c1x2 + c2x22

This can be re-written as a linear system: 1 x0 x

20

1 x1 x21

1 x2 x22

c0c1

c2

=

y0y1

y2

Generalize for n + 1 points: (x0, y0), (x1, y1), . . . (xn, yn)

1 x0 x20 xn01 x1 x

21 x

n1

... ... ... ...

1 xn x2n x

nn

A

c0c1

c2...

cn

c

=

y0y1

y2...

yn

y

. . . this is a linear system for the coefficients c.

A is an (n + 1) (n + 1) matrix, and is called a Vandermondematrix.



6/13


Vandermonde Example

Given the following gasoline price data (obviously out of date!):

Year 1986 1988 1990 1992 1994 1996

Price (cents) 66.7 66.1 69.3 70.7 68.8 72.1

Use polynomial interpolation to approximate the gas price in 1991.

Problems:

The Vandermonde matrix is usually ill-conditioned, which can

cause large errors in the polynomial coefficients.

Solving a dense matrix system requires O(n3

) floating pointoperations, which is expensive.

We want a cheaper and more accurate method!



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Lagrange Interpolation

Again, start with n + 1 points satisfying yi = f(xi)

(x0, y0), (x1, y1), . . . (xn, yn)

Idea: Construct a simple polynomial that is equal to 1 at x0and equal to 0 at every other point:

l0(x) :1 0 0 0| | | |

x0 x1 x2 xn

The following does the trick:

L0(x) =(x x1)

(x0 x1)

(x x2)

(x0 x2)

(x xn)

(x0 xn)=

ni=1

x xi

x0 xi

Construct similar polynomials for the other points

Lj(x) =n

i=0i=jx xi

xj xiwhich satisfy Lj(xi) =

1 if i = j

0 if i = j

Lj(x) is called the jth Lagrange polynomial.

We can now write the interpolating polynomial for f(x) as

Pn(x) = y0L0(x) + y1L1(x) + + ynLn(x)

=n

j=0

yjLj(x)

so that

Pn(xk) = y0L0(xk) + y1L1(xk) + + ynLn(xk)

= ykLk(xk) = yk



8/13


Lagrange Interpolation Example

Back to the data from our first example:

x y = f(x)

0.0 0.00000.5 1.5163

1.0 3.6788

2.0 5.4134

2.5 5.1303

3.0 4.4808

4.0 2.9305

x y = f(x)

5.0 1.68456.0 0.8924

7.0 0.4468

8.0 0.2147

9.0 0.1000

10.0 0.0454

Find the quadratic interpolating polynomial P2(x) on [1, 2.5]:

First set up the Lagrange polynomials:

L0(x) =(x2)(12)

(x2.5)(12.5)

= (x2)(x2.5)1.5

(leave unexpanded!!!)

L1(x) =(x1)(21)

(x2.5)(22.5)

= (x1)(x2.5)0.5

L2(x) =(x1)

(2.51)(x2)

(2.52)= (x1)(x2)

0.75

Then construct the interpolating polynomial:P2(x) =

3.67881.5

(x 2)(x 2.5) + 5.41340.5

(x 1)(x 2.5)

+ 5.13030.75

(x 1)(x 2)

= 2.4525 (x 2)(x 2.5) 10.8268 (x 1)(x 2.

+ 6.8404(x 1)(x 2)

and P2(1.75) 5.26735

Never expand the Lagrange form . . . except to compare:

P2(x) = 1.1235 + 6.3362x 1.5339x2

(same as before . . . UNIQUE!!)



9/13


Pros and Cons of the Lagrange Form

Advantages of the Lagrange form of the interpolating polynomial:

Polynomial can be written down easily great for manual

computation!

Cost of O(n2) is much better Vandermonde was O(n3).

Disadvantages:

If we add another point, (xn+1, yn+1), then we have to start

over from scratch none of the Lk(x) can be reused.

No simple way to estimate error were stuck using the errorterm En(x) from the Theorem on page 3:

En(x) =f(n+1)(c)

(n + 1)!(x x0)(x x1) (x xn)

for which f(n+1)(x) is typically not known!



10/13


Newton Divided Differences

Another way to construct the interpolating polynomial, Pn(x).

Idea: Build up one point and one degree at a time.

Start at degree 0, where P0(x) interpolates a single point (x0, y0)

P0(x) = y0

Move to degree 1, by adding a term a1(x x0):

P1(x) = y0 +y1 y0

x1

x0

(x x0)

This pattern can be continued to higher degree:

P0(x) = a0

P1(x) = a0 + a1(x x0)

P2(x) = a0 + a1(x x0) + a2(x x0)(x x1)

Pn(x) = a0 + a1(x x0) + a2(x x0)(x x1)+ + an(x x0)(x x1) (x xn1)

=n

j=0

aj

j1i=0

(x xi)

We now substitute Pn(xk):

y0 = Pn(x0) = a0 = a0 = y0

y1 = Pn(x1) = a0 + a1(x1 x0) = a1 =y1y0x1x0

which is a divided difference.



11/13


The next divided difference comes from

y2 = Pn(x2) = a0 + a1(x2 x0) + a2(x2 x0)(x2 x1)

= a2 =

y2y1x2x1

y1y0x1x0

x2 x0

It helps to introduce some notation:

f[xi] = yi

f[xi, xi+1] =f[xi+1] f[xi]

xi+1 xi

f[xi, xi+1, xi+2] =f[xi+1, xi+2] f[xi, xi+1]

xi+2 xi...

f[xi, . . . , xi+k] =f[xi, . . . , xi+k1] f[xi+1, . . . , xi+k]

xi+k xi

Then, the coefficient ak is given by

ak = f[x0, x1, . . . , xk]

and is called the kth divided difference.

The calculations are much simplerwhen organized as a table:

x y a1 a2 a3 a4x0 f[x0]

f[x0, x1]x1 f[x1] f[x0, x1, x2]

f[x1, x2] f[x0, x1, x2, x3]

x2 f[x2] f[x1, x2, x3]. . .

f[x2, x3] f[x1, x2, x3, x4]

x3 f[x3] f[x2, x3, x4]. . .

f[x3, x4] f[x2, x3, x4, x5]x4 f[x4] f[x3, x4, x5]

f[x4, x5]x5 f[x5]

...



12/13


Newton Divided Differences Example

Use data from our previous example to construct the table.

Take the points x = 0.0, 0.5, 1.0, 2.0, 2.5, 3.0.

x y a1 a2 a3 a40.0 0.0000

3.03270.5 1.5163 1.2923

4.3249 1.50961.0 3.6788 1.7269 0.6424

1.7346 0.09652.0 5.4134 1.5339 0.12160.5662 0.4006

2.5 5.1303 0.7328

1.29903.0 4.4808

The coefficients ak can be read directly off the diagonals.

Starting at the point x = 0.0:

P0(x) = 0.0000

P1(x) = 0.0000 + 3.0327(x 0.0)

P2(x) = 3.0327(x 0.0) + 1.2923(x 0.0)(x 0.5)

P3(x) = P2(x) 1.5096(x 0.0)(x 0.5)(x 1.0)

etc.

Starting at the pointx

= 1.0 (as in the earlier examples):

P0(x) = 3.6788

P1(x) = 3.6788 + 1.7346(x 1.0)

P2(x) = 3.6788 + 1.7346(x 1.0) 1.5339(x 1.0)(x 2.0)

etc.

. . . exactly as before!



13/13


Advantages of Newton Divided Differences:

Can extract the interpolating polynomial for any interval and

any degree desired.

Cost is half that of Lagrange polynomial.

Easy access to an error estimate the error term can be writ-

ten as

En(x) = f[x0, x1, . . . , xn+1]n

i=0

(x xi)

(see Exercise 20, page 129). In practice, error can be esti-

mated by adding one more point and computing one extra

row in the table.


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interpolation.pdf