1. Computational Methods in Physics PHYS 3437
Dr Rob Thacker
Dept of Astronomy & Physics (MM-301C)

2. Today’s Lecture Interpolation & Approximation I Overview of ideas
Interpolation vs fitting
Polynomial interpolation
Lagrange interpolation
Hermite interpolation
Cubic splines (introduction)

3. General idea of interpolation
Suppose we are given data at discrete points only, and we wish to estimate values between these known points x1 x2 x3 x4 x5
f(x) X

4. Difference between interpolation & fitting
Interpolation passes through each datum
Fitting seeks to produce a curve that approximates the data x1 x2 x3 x4 x5
f(x)

5. Global versus piecewise interpolation
We can fit a single polynomial of degree n to n+1 data points
Alternatively we can use a series of polynomials to link points together – frequently called splines
Splines will usually be cubic polynomials – I’ve used linear here to emphasize the piecewise nature x1 x2 x3 x4 x5
f(x)

6. Lagrange interpolation
To fit a polynomial of degree n we need n+1 points
Consider first interpolating between two points, f(x2),f(x3) using a function F(x)
The Taylor expansion about a point x gives
f(x2)=F(x)+(x2-x)F’(x)+…
f(x3)=F(x)+(x3-x)F’(x)+… (I’ve highlighted what we know)
If we use a polynomial of degree 1, p(x), then the Taylor series truncates after p’(x)
f(x2)=p(x)+(x2-x)p’(x) (1)
f(x3)=p(x)+(x3-x)p’(x) (2)

7. Eliminate p’(x)
After a short derivation (ex) we find that eqns (1) and (2) give
On varying the expansion point x, but keeping fixed end points x2,x3 this corresponds to a straightline – as expected (check the following for yourself):
If x=x3 => p(x)=f(x3)
If x=x2 => p(x)=f(x2)
Linear fits aren’t that helpful though, and bringing in more points allows us to up the order of the interpolation

8. Quadratic fit using three points
If we again Taylor expand put assume the function is a polymial, p(x) of degree 2, then
f(x1)=p(x)+(x1-x)p’(x)+(x1-x)2p’’(x)/ (3)
f(x2)=p(x)+(x2-x)p’(x)+(x2-x)2p’’(x)/ (4)
f(x3)=p(x)+(x3-x)p’(x)+(x3-x)2p’’(x)/ (5)
With a bit more algebra (fairly lengthy) we can eliminate both p’(x) and p’’(x) using (3),(4),(5) to get
You should see a pattern emerging here…

9. General (Lagrange) Polynomial fit
If we have n points, a polynomial of degree n-1 can be constructed using the formula
This is called the interpolation polynomial in Lagrange form, but it is frequently referred to as the Lagrange polynomial

10. Properties of the lj,n(x)
The lj,n(x) are clearly polynomials, and a called basis polynomials since they sum to find the interpolating polynomial
If we let x=xi in the formula (so that we are picking one of the interpolation points) then

11. Problems with polynomial interpolation
As the number of points increases so does the degree of the polynomial
This implies many turns and a lot of “structure” between interpolation points in high order polynomials
High order polynomials will have a lot of “wiggles” and if points do not change smoothly they can “overshoot” in between datums
Cubic fits tend to be about optimal
4 data points can surround x symmetrically (2 on each side)
Increasing the number of points adds more data from further away from a given interpolation point
This doesn’t necessarily improve the accuracy of the fit locally

12. Example of “wiggles” Piecewise linear fit 6th order polynomial fit
From

13. Segue - numerically implementing the general formula
Implementing the general formula is not exactly trivial
It helps to notice that the lj,n(x) can be written as a product
One can then think about using a loop to do the multiplication of the terms in the product
However, the best way to calculate large interpolation polynomials is to use Neville’s Algorithm

14. Segue - Neville’s Algorithm
We won’t go into the details, however you can take a look (if you are interested) at Numerical Recipes section 3.1
Neville’s algorithm is better because it builds up the interpolation polynomial in a recursive fashion
It uses a similar table idea to building up coefficients in a binomial expansion
The employed recursion cuts down the total number of operations required and helps reduce concerns about rounding error

15. Hermite Interpolation
If we have both the function value, and the derivative we can fully constrain a higher order polynomial through two points
f(x)
f(x2) & f’(x2)
f(x1) & f’(x1) x1 x x2
(x1<x<x2)

16. Interpolating a cubic to two points
Given the general form for a cubic
So we have 4 equations
with 4 unknowns to solve.

17. Solve for p(x)
While again somewhat lengthy, we can solve for p(x) in terms of the known values (subscript denotes for x1,x2)
One important property of Hermite interpolation is that if we fit a different polynomial p2,3(x) between x2 & x3 then it will have the same derivative at x2 as p1,2(x2) i.e. p’1,2(x2)=p’2,3(x2)
Interpolation between successive pairs of points is continuous
The derivatives are also continuous
Hermite interpolation is thus considered to be smooth

18. Diagram of two Hermite interpolation polynomials
Second polynomial fit
p2,3(x)
First polynomial fit
p1,2(x)
f(x)
f(x3) & f’(x3)
f(x1) & f’(x1)
f(x2) & f’(x2)
Derivatives match at x2 x1 x2 x3

19. Advantages of Hermite interpolation over Lagrange interpolation
A series of H.I. polynomials will be continuous at the datums, as will the derivatives (i.e. smooth)
While a series of L.I. polynomials is continuous at the datums the derivatives will not be continuous
H.I. can fit a cubic to `local’ data (i.e. x1<x<x2)
L.I. requires four points to fit a cubic (x1<x2<x<x3<x4)
However, even if we don’t have derivatives it would be nice to fit a series of
polynomials with continuous slopes at each datum

20. Cubic Splines
Let’s examine how we can create a series of cubic polynomials with equal derivatives at the datums
Choose x, such that xj<x<xj+1 where j=1,…,n
Around x, we may write p(x) as

21. Next consider derivatives
We just derived two equations (3) & (2) for the value of the cubic fit at xj and xj+1
Taking the first & second derivative of p(x) we find
If we again equate at xj and xj+1 but using 2nd deriv:

22. Next find the cj
We’ve just shown

23. Substitute back to get p(x)
We now substitute (2),(4),(5) and (6) back into our first equation for p(x) to get
To get any further we now need some information about p’’j+1 and p’’j

24. Setting p´´j+1 and p´´j f(x) f(x3) f(x1) f(x2) x1 x2 x3
We have no information about the derivative of f(x) though
We just have the series of f(x) values
What requirements can we impose?
Recall we want the derivatives of neighbouring polynomials to match at the datums (xj)
f(x)
f(x3)
f(x1)
f(x2)
Derivatives match at x2 x1 x2 x3

25. Apply continuity in the derivative
We now have to consider adjacent polynomials
Label:
pj,j+1 to be the fit between xj and xj+1
pj-1,j to be the fit between xj-1 and xj
Continuity of the derivative implies that p´j,j+1 and p´j-1,j should be equal at xj, the general p’ forms are:

26. Equate at xj
Setting p´j,j+1(xj)=p´j-1,j(xj) the first few terms of (8) vanish, while terms in (x-xj-1) in (9) cancel with hj-1:
Where j=2,3,…,n-1

27. Finding the p´´j values
We now have a system of n-2 equations
Recall limits of j=2,3,…,n-1
There are n unknown values though
The p’’j values range through j=1,…,n
To have enough information to solve for all the values we have to provide information about the end points – two cases
Specify derivatives at end points
“Clamped spline”
Set p’’1=0 and p’’n=0
“Natural spline”

28. Summary
Interpolation shouldn’t be confused with fitting, although the terms are frequently used interchangeably
Lagrange interpolation will fit an order n polynomial to n+1 data points
Hermite interpolation uses derivatives to solve for a cubic fit with only 2 datums
Cubic spline interpolation gives a smooth piecewise interpolation scheme but requires conditions to be set for the end points

29. Next Lecture Matrix forms for spline fits Tridiagonal solvers
Clamped spline
Natural spline
Tridiagonal solvers