scientific computing interpolation – divided differences efficiency and error analysis

Scientific Computing

Interpolation – Divided Differences Efficiency and Error Analysis

How is Interpolation used?

• Here is a table of values for the temperature at a given depth for a lake. We want a good estimate of the temperature at a depth of 7.5 meters.

Temperature Depth

T (oC) z (m)19.1 019.1 -119 -2

18.8 -318.7 -418.3 -518.2 -617.6 -711.7 -89.9 -99.1 -10

Newton’s Nested Formula

• Recall: Newton’s method is iterative with Pk+1 (x) = Pk (x) + ck (x – x0)(x – x1) … (x – xk)

• Now, Pk (x) satisfies a similar formula:

Pk (x) = Pk-1 (x) + ck-1 (x – x0)(x – x1) … (x – xk-1)

• So, Pk+1 (x) = Pk-1 (x) + [ck-1 (x – x0)…(x – xk-1)] +

+ [ck (x – x0)…(x – xk)]

• Thus, Pn (x) = c0 + [c1 (x – x0)] + … + [cn (x – x0 )…(x – xn )]

Newton Nested Formula

• Pn (x) = c0 + [c1 (x – x0)] + … + [cn (x – x0 )…(x – xn )]

• We can write this in a compact form as

• Here we define the product for k =0 to be 1, so the first term is c0 , as it should be.

• We can rewrite this formula as (verify this!!) Pn (x) = c0 + (x – x0)[c1 + (x – x1)[c2 + (x – x2)[…]]]

pn (x) ck (x x j )0jk

k0

n

Newton Nested Formula

• Pn (x) = c0 + (x – x0)[c1 + (x – x1)[c2 + (x – x2)[…]]]

• This allows a better way of calculating Pn (t) for a value of t (or x):

• Then, vn = Pn (x) • This is called Newton’s Nested formula for Pn

100

1222

0111

0

)(

)(

)(

nn

nn

nn

n

vxxcv

vxxcv

vxxcv

cv

Efficiency

• Newton’s Nested formula Requires 2n additions, n multiplications O(n) • Lagrange Interpolation Requires at least n2 additions O(n2)

100

1222

0111

0

)(

)(

)(

nn

nn

nn

n

vxxcv

vxxcv

vxxcv

cv

P(x) x x jxk x jjk

k

yk

Newton Divided Differences

• vn = Pn (x)

• This is efficient, but is there a fast way to calculate the ck ‘s?

100

1222

0111

0

)(

)(

)(

nn

nn

nn

n

vxxcv

vxxcv

vxxcv

cv


• Pn (x) = c0 + c1 (x – x0) + … + cn (x – x0 )…(x – xn )]

1) Substituting x0 for x and y0 for y=Pn(x), we get

c0 = y0=f[x0]

2) Now, substituting x1 for x and y1 for y, we get

3) Substitute once more (x2 ,y2) to get

)],[)( 10001

01101101 xxfDy

xx

yycxxcyy

)())((

)()(

))(()(

02

12

02

01

01

12

02

1202

0201

0102

2

120220201

0102

xx

xx

xx

xx

yy

xx

yy

xxxx

xxxx

yyyy

c

xxxxcxxxx

yyyy


3) (cont’d) Simplifying, we get

So, Pn (x) = f[x] + f[x0, x1](x – x0) + f[x0, x1, x2](x – x0)(x -x1) + …

+ cn (x – x0 )…(x – xn )]

],,[)(

)()(

1

)(

)()(

)(

21002

02

01

02

01

01

12

12

02

01

01

12

01

01

01

12

02

02

12

01

01

01

12

02

02

12

0112

01

01

12

02

02

12

02

01

01

12

02

2

xxxfyDxx

DyDy

xx

xx

yy

xx

yy

xx

xx

yy

xx

xx

xx

yy

xx

yy

xx

xx

xx

xx

yy

xx

yy

xx

xx

xxxx

xx

yy

xx

yy

xx

xx

xx

xx

yy

xx

yy

c


• Definition (Divided Differences). For a given collection of data { (xi ,f(xi) ) } a kth order divided difference is a function of k + 1 (not necessarily distinct) data values written as f[xi , xi+1 , … , xi+k].

0th Order: f[xi ] = f(xi) = yi

kth Order:


• Theorem For Newton’s Interpolation formula Pn (x) = c0 + [c1 (x – x0)] + … + [cn (x – x0 )…(x – xn )]

the coefficients ck can be calculated as

ck = f[x0 , x1 , … , xk]


• Algorithm for Calculating Divided Differences: Calculate the differences in “pyramid” fashion, until you reach the last one.


• Example: Data (1,3), (1/2, -10), (3,2)


• Example: Data (1,3), (1/2, -10), (3,2)

Top row of the pyramid contains coefficients, c0 = 3, c1 = 26 , c2 = -53/5Recall: Earlier work gave p(x) = 3+ 26 (x-1) – (53/5) (x-1)(x-1/2)

Matlab - Newton Divided Differences

• Notes: We want to store the values in the “pyramid”. These can be stored in a matrix. We let F(i,k) = kth divided difference

• Differences are indexed from 0 to n, so F will have indexes from 1 to n+1.

• Vectors: x[], y[] for initial data, y = f(x).


• Notes: F(i,k) = kth divided difference

• Get F(i,k+1) = (F(i+1,k)-F(i,k))/(x(i+k)-x(i))


function F = div_diff(x, y)% div_diff function computes all divided differences for % the data stored in x and y = f(x)n = length(x);m = length(y);if m~=n; error('x and y must be same size');else F = zeros(n, n); for i = 1:n % define 0th divide difference F(i,1) = y(i); end for k = 1:n-1 % Get kth divided difference for i = 1:n-k F(i,k+1) = (F(i+1,k)-F(i,k))/(x(i+k)-x(i)); end end


• Example: x=[1 0.5 3]‘; y=[3 -10 2]'; div_diff(x,y) ans = 3.0000 26.0000 -10.6000 -10.0000 4.8000 0 2.0000 0 0

• Top row (F(1,*)) holds coefficients ck .

Matlab - Newton Nested Formula

• Need loop running “Backwards” • For evaluation, we replace x by a value u. 100

1222

0111

0

)(

)(

)(

nn

nn

nn

n

vxxcv

vxxcv

vxxcv

cv


function v = newtonInterp(x,y,u)% v = newtonInterp(x,y,u) computes the Newton method % interpolation value p(u) for the given data x and yn = length(x);m = length(y);if m~=n; error('x and y must be same size');else F = div_diff(x,y); % Compute Newton polynomial using nested format v = F(1,n); % v0 = cn for i = n-1:-1:1 v = F(1,i) + (u-x(i))*v; % nested formula endend


• Test:>> newtonInterp(x,y,1)ans = 3>> newtonInterp(x,y,0.5)ans = -10>> newtonInterp(x,y,3)ans = 2

Class Exercise

• Here is a table of values for the temperature at a given depth for a lake. Find a good estimate of the temperature at a depth of 7.5 meters. Use a cubic interpolating polynomial to help solve this problem.

Temperature Depth

T (oC) z (m)19.1 019.1 -119 -2

18.8 -318.7 -418.3 -518.2 -617.6 -711.7 -89.9 -99.1 -10

Error Analysis

1) If we want to interpolate a given f(x) at n+1 nodes, how should we pick nodes?

2) How accurate can we make a polynomial interpolant over an interval?

scientific computing interpolation – divided differences efficiency and error analysis

Documents