Spline Interpolation Method - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Spline Interpolation Method

What is Interpolation?

Given (x

,y

), (x

,y

), …… (x

n

,y

n

), find the

value of ‘y’ at a value of ‘x’ that is not given.

Why Splines?

2 1 25

1 ( )

x

f x

=

Table : Six equidistantly spaced points in [-1, 1]

Figure : 5

th order polynomial vs. exact function

x (^) 2 1 25

1

x

y

=

• 1.0 0.
• 0.6 0.
• 0.2 0.

0.2 0.

0.6 0.

1.0 0.

Why Splines?

Figure : Higher order polynomial interpolation is a bad idea

-0.

-0.

0

-1 -0.5 0 0.5 1

x

y

19th Order Polynomial f (x) 5th Order Polynomial

Linear Interpolation (contd)

( ),

( ) ( ) ( ) ( ) 0 1 0

1 0 0 x x x x

f x f x f x f x − −

− = + x (^) 0 ≤ x ≤ x 1

( ),

( ) ( ) ( ) 1 2 1

2 1 1 x x x x

f x f x f x − −

− = + x 1 (^) ≤ x ≤ x 2

.

.

.

( ),

( ) ( ) ( ) 1 1

1 1 − −

− − − −

− = + n n n

n n n x x x x

f x f x f x x (^) n − 1 ≤ x ≤ xn

Note the terms of

1

( ) ( 1 )

−

− −

−

i i

i i x x

f x f x

in the above function are simply slopes between xi − 1 and x i.

Example

The upward velocity of a rocket is given as a function

of time in Table 1. Find the velocity at t=16 seconds

using linear splines.

Table Velocity as a

function of time

Figure. Velocity vs. time data

for the rocket example

(s) (m/s)

t v ( t )

Quadratic Interpolation

Given ( x (^) 0 , y 0 ) (, x 1 , y 1 ),......, ( xn − 1 , yn − 1 ) (, xn , yn ), fit quadratic splines through the data. The splines

are given by

( ) 1 1 ,

2 f x = a 1 x + bx + c x (^) 0 ≤ x ≤ x 1

2 2 ,

2 = a (^) 2 x + b x + c x 1 (^) ≤ x ≤ x 2

.

.

.

,

2 = a (^) n x + bnx + c n x (^) n − 1 ≤ x ≤ xn

Find a i , bi , ci , i =1, 2, …, n

Quadratic Interpolation (contd)

Each quadratic spline goes through two consecutive data points

1 0 1 (^0 )

2 a 1 (^) x 0 + bx + c = f x

1 1 1 (^1 )

2 a 1 (^) x 1 + bx + c = f x.

.

.

1 (^1 )

2 ai xi − 1 + bixi − + ci = f xi −

( )

2 a (^) i xi + bixi + ci = f x i.

.

.

1 (^1 )

2 an xn − 1 + bnxn − + cn = f xn −

( )

2 a (^) n xn + bnxn + cn = f x n

This condition gives 2n equations

Quadratic Splines (contd)

Similarly at the other interior points,

2 a 2 x 2 + b 2 − 2 a 3 x 2 − b 3 = 0

.

.

.

2 ai xi + bi − 2 ai + 1 xi − bi + 1 = 0

.

.

.

2 an − 1 xn − 1 + bn − 1 − 2 anxn − 1 − bn = 0

We have (n-1) such equations. The total number of equations is ( 2 n ) + ( n − 1 )=( 3 n − 1 ).

We can assume that the first spline is linear, that is a 1 = 0

Quadratic Splines (contd)

This gives us ‘3n’ equations and ‘3n’ unknowns. Once we find the ‘3n’ constants,

we can find the function at any value of ‘x’ using the splines,

( ) 1 1 ,

2 f x = a 1 x + b x + c x 0 (^) ≤ x ≤ x 1

2 2 ,

2 = a (^) 2 x + b x + c x 1 (^) ≤ x ≤ x 2

.

.

.

,

2 = a (^) n x + bnx + c n x (^) n − 1 ≤ x ≤ xn

Solution

1 1

2

1

v t = a t + b t + c 0 ≤ t ≤ 10

, 2 2

2

2

= a t + b t + c 10 ≤ t ≤ 15

, 3 3

2

3

= a t + b t + c 15 ≤ t ≤ 20

, 4 4

2

4

= a t + b t + c 20 ≤ t ≤ 22. 5

, 5 5

2

5

= a t + b t + c 22. 5 ≤ t ≤ 30

Let us set up the equations

Each Spline Goes Through Two

Consecutive Data Points

( ) , 1 1

2

1

v t = a t + b t + c 0 ≤ t ≤ 10

( 0 ) ( 0 ) 0 1 1

2

1

a + b + c =

( 10 ) ( 10 ) 227. 04 1 1

2

1

a + b + c =

Derivatives are Continuous at

Interior Data Points

v t = a t + b t + c^0 ≤^ t ≤^10

2 2

2

2

= a t + b t + c 10 ≤ t ≤ 15

t t

a t b t c

dt

d

a t b t c

dt

d

t t

a t b a t b

2 a 10 + b = 2 a 10 + b

a + b − a − b =

Derivatives are continuous at

Interior Data Points

2 ( 10 ) 2 ( 10 ) 0 1 1 2 2

a + b − a − b =

2 ( 15 ) 2 ( 15 ) 0 2 2 3 3

a + b − a − b =

2 ( 20 ) 2 ( 20 ) 0 3 3 4 4

a + b − a − b =

2 ( 22. 5 ) 2 ( 22. 5 ) 0 4 4 5 5

a + b − a − b =

At t=

At t=

At t=

At t=22.