Interpolation

• In interpolation or extrapolation we usually want to do the following

- We have data points xi yi , i = 12} N

- We want to know the y value at xxz i

- In interpolation x1 xxN and in extrapolation xx 1 or xx! N - Extrapolation is dangerous; it is used e.g. in solving differential equations.

- Data set may have noise: the interpolate should go smoothly through the data set not necessarily through all points.

- One application is approximating (special) functions - In this case we have an infinite number of points available.

- In some cases interpolation is done by using a few points in the neighborhood of x . - This may result in noncontinuous derivative of the interpolate. - In spline interpolation one condition is that also the derivative is continuous.

- In polynomial interpolation one is not particularly interested in the polynomial coefficients only in its values. - Calculating coefficients is rather error prone.

Scientific computing III 2011: 6. Interpolation 1

Interpolation

• Interpolation vs. curve fitting:

Scientific computing III 2011: 6. Interpolation 2 Interpolation

• Degree of interpolation is (number of points used)-1. - Below an example of interpolation of a smooth function

original function

low degree interpolation

high degree interpolation

- When the original function has sharp corners an interpolation polynomial with a lower degree may work better

Scientific computing III 2011: 6. Interpolation 3

Interpolation: polynomials

• We have a data set xi yi , yi = fx()i , i = 1} N

- We have to find a polynomial P N – 1 x that fulfills the condition

PN – 1()xi = yi , i = 1} N

- It is easy to show that the polynomial is at most of degree N – 1 and it is unique if all x i are different.

- A straightforward way to determine the coefficients is to use the methods we have already learned

- Let the polynomial be of the form

2 N–1 yc=1+c2xc+3x ++}cNx

- From the above condition we get a group of linear equations

­ 2 N–1 °c1+c2x1+c3x1 ++}cNx1 =y1 ° °c+cx+cx2 ++}cxN–1=y ®12232 N2 2 ° } ° °c+cx+cx2 ++}cxN–1=y ¯12N3N NN N

Scientific computing III 2011: 6. Interpolation 4 Interpolation: polynomials

• Theorem: existence of interpolating polynomial:

If points x1 x2 } xN are distinct, then for arbitrary real values y1 y2 } yN there is a unique polynomial P of

degree d N – 1 such that Px i = yi for 1 d i d N .

- Proof by induction:

Suppose we have already a polynomial P that reproduces a part of the data set: Px i = yi , 1 ddik ,

(For example: this can be a constant polynomial for data point 1: Px = y1 .)

Then we add another term to P so that it will go through the point xk + 1 yk + 1 :

Qx = Px + cxx – 1 xx– 2 } xx– k

Q reproduces data points 12 } k because P does and the added term is zero for all these points.

Now we adjust constant c so that Q reproduces the data point k + 1 :

Qx k + 1 = Px k + 1 + cx k + 1 – x1 xk + 1 – x2 } xk + 1 – xk = yk + 1 .

From this equation we can solve c because all x i are distinct.

QED.

Scientific computing III 2011: 6. Interpolation 5

Interpolation: polynomial

- In matrix form:

1 x x2 } xN – 1 1 1 1 c1 y1 2 N – 1 1 x2 x2 } x2 c2 y2 2 N – 1 ˜ c = y 1 x3 x3 } x3 3 3 }}}} } } } 2 N–1 cN yN 1 xNxN}xN

- This square matrix has a name of its own: Vandermonde matrix and its a little bad behaving:

>> v=1:5 >> v=1:8 v = v = 1 2 3 4 5 1 2 3 4 5 6 7 8 >> m=vander(v) >> m=vander(v) m = m = 1 1 1 1 1 1 1 1 1 1 1 ... 16 8 4 2 1 128 64 32 16 8 4 ... 81 27 9 3 1 2187 729 243 81 27 9 ... 256 64 16 4 1 16384 4096 1024 256 64 16 ... 625 125 25 5 1 78125 15625 3125 625 125 25 ... >> rcond(m) 279936 46656 7776 1296 216 36 ... ans = 823543 117649 16807 2401 343 49 ... 2.2699e-05 2097152 262144 32768 4096 512 64 ... >> rcond(m) ans = 6.0171e-10

Scientific computing III 2011: 6. Interpolation 6 Interpolation: polynomial

- A better way is to calculate P N – 1 by using so called Lagrange’s polynomials.

- Let the polynomials l1 l2 } lN be defined as

N§·xx– l ()x = ¨¸------j, i=1}N i –©¹x–x j=1i j jiz

- These functions have the property

li()xj= Gij

- Now we can write P N – 1 x as

N P ()x = fx()l()x N – 1 ¦ i i i=1

- It is easy to check that P N–1 x goes through all the data points xiyi .

- Because functions l i have degree less than N it follows that P N–1 also has degree less than N .

Scientific computing III 2011: 6. Interpolation 7

Interpolation: polynomial

- Example:

x : 13e 14e 1 43e y : 2 –1 7 2

1 4 §·x– - -- x–1§·x– - -- ©¹4 ©¹3 93 69 l x = ------= 1 8 x 3 – ------x 2 + ------x – 6 1 11 1 14 2 2 §·- -- – - -- §·---1– §·- -- – - -- ©¹34 ©¹3©¹33 1 4 §·x– - -- x–1§·x– - -- ©¹3 ©¹3 192 512 1216 256 l x = ------= – ------x 3 + ------x 2 – ------x + ------2 11 1 14 13 13 39 39 §·- -- – - -- §·---1– §·- -- – - -- ©¹43 ©¹4©¹43 1 1 4 §·x– - -- §·x– - -- §·x– - -- ©¹3©¹4©¹3 23 31 2 l x = ------= – 6 x 3 + ------x 2 – ------x + - -- 3 1 1 4 2 6 3 §·1– - -- §·1– - -- §·1– - -- ©¹3©¹4©¹3 1 1 §·x– - -- §·x– - -- x–1 ©¹3©¹4 36 57 24 3 l x = ------= ------x 3 – ------x 2 + ------x – ------4 1 41 4 13 13 13 13 §·-4 -- – - -- §·- -- – - -- §·---1– ©¹33 ©¹34 ©¹3 186 1577 5281 560 P x = 2l x–l x+7l x+2l x = ------x 3 – ------x 2 + ------x – ------3 1 2 3 4 13 26 78 39

Scientific computing III 2011: 6. Interpolation 8 Interpolation: polynomials

- So we end up with the interpolation polynomial (let’s leave the subscript out)

xx–2 xx–3 } xx–N Px()= ------y 1 x1–x2 x1–x3 } x1 – xN xx–1 xx–3 } xx–N + ------y 2 x2–x1 x2–x3 } x2 – xN + }

xx–1 xx–2 } xx–N–1 + ------y N xN–x1 xN–x2 } xN – xN – 1

- The error estimation of the above polynomial can be given as follows:

- Let x 1 x 2 } x N be distinct numbers in >@ab  and let’s assume that f has N continuous derivatives in >@ab - Then for each x >@ ab  [ x  ab so that

N f N [ x fx =Px +------xx– (1) N!– i i=1

Scientific computing III 2011: 6. Interpolation 9

Interpolation: polynomials

- Proof:

- For xx= k , k = 12 } N fx k = Px k and any [ xk  ab fulfills (1)

- For x k z x we define function gt as tx– gt = ft –Pt – >@fx –Px N ------i(2) – i = 1 xx–i

- Since f has N continuous derivatives and P has all derivatives continuous and xx z k o gt has N continuous deriv- atives in >@ab

- For tx= k Generalized Rolle’s theorem: N x–x gx = fx –Px – >@fx –Px ------ki= 0 Assume k k k – 1. fx continuous on >@ab , xx–i i = 1 2. derivatives f 1 }x f N x exist in ab , 3. x x x ab, - For tx= 01 } N  >@ 4. fx j = 0 , for j = 01} N . N xx– Then , , such that N . gx = fx –Px – >@fx –Px ------i= 0 c acb f c = 0 – xx– i = 1 i

- Thus g vanishes at N + 1 points xx 1 x 2 } x N in >@ab . - Generalized Rolle’s theorem says that  [[ { x in ab for which g N [ =0 . - From (2) we get dN­½ tx– 0=g N [ =f N [–P N [–>@fx –Px ------N ------i (3) N®¾– i = 1 dt ¯¿ xx–i t=[

Scientific computing III 2011: 6. Interpolation 10 Interpolation: polynomials

- Now P is at most of degree N – 1 o P N { 0 - The product term is a polynomial of degree N o

N N –1 N tx– dN­½ tx– N! ------i= xx– tN+Ot N–1 o ------i = ------– – i N®¾– N xx–i dt ¯¿ xx–i xx– i = 1 i=1 i = 1 –i=1 i - (3) now becomes N! 0 = f N [– >@fx –Px ------N xx– –i=1 i - And solving for fx we get f N [ fx =Px +------N xx– N! –i=1 i

QED.

- Note the analogy between the error formula of the Taylor series: f N [ ------xx– N N! 0 and f N ------[ xx– xx– } xx– N! 1 2 N

Scientific computing III 2011: 6. Interpolation 11

Interpolation: polynomials

- Example: - Prepare a table for function fx =ex , x>@01 . - Precision d decimals, step size h - What step size is needed for linear interpolation to give absolute error no more than 10–6 ?

- Let x  >@01 and xj d x d xj + 1 - Error is now f 2 f 2 f 2 fx –Px d------[ xx– xx– =------[ xx– xx– =------[ x– jh xj– +1h 2! j j+1 2 j j+1 2 1 fx –Px d---max f 2 [ max x– jh xj– +1h 2 [ >@01 xj ddxxj+1 Maximum of x– jh xj– +1h is at xj= +12eh with value h2e4

eh2 o fx –Px d ------8

- We want the error to be less than 10–6 : eh2 8 –3 ------d 1 0 – 6 ŸŸh 2 d ---10–6 h 1.72u10 8 e

- So we could choose h=0.001

Scientific computing III 2011: 6. Interpolation 12 Interpolation: polynomials

- As a figure:

Scientific computing III 2011: 6. Interpolation 13

Interpolation: polynomials

- In principle one could calculate the interpolating values from the above equation we obtained:

xx–2 xx–3 } xx–N Px()= ------y 1 x1–x2 x1–x3 } x1 – xN xx–1 xx–3 } xx–N + ------y 2 x2–x1 x2–x3 } x2 – xN + }

xx–1 xx–2 } xx–N–1 + ------y N xN–x1 xN–x2 } xN – xN – 1

- However, there are more efficient ways to do that: so called Neville’s algorithm:

- Let P 1 be a zero-degree polynomial going throught the first point x 1  y 1 ; i.e. P 1 = y 1 .

- In the same way define P 2 P 3 } P N

- Let P 12 be a first-degree polynomial going through points x 1 y1 and x2 y2 ,

- In the same way define P23P34 } P N + 1 N

- Going further we can define polynomials with higher degrees until we get P 123}N which is what we want.

Scientific computing III 2011: 6. Interpolation 14 Interpolation: polynomials

- Different P ‘s can be written as an array with parents and children; for N = 3 :

x1 y1 = P1

P12

x2 y2 = P2 P123

P23

x3 y3 = P3

- Neville’s algorithm fills the above array from left to right one column at a time.

- The recursion formula for the P ‘s is

Pii +1 } im+ xx– P + x–xP = ------im+ ii +1 } im+ –1 i i+1 } im+ xi–xim+

- This recursion works because the parents have the same values at points x i + 1 } x im+ –1

Scientific computing III 2011: 6. Interpolation 15

Interpolation: polynomials

- Neville’s algorithm is based on Aitken’s lemma:

Let P 1n x be a Lagrange’s polynomial with degree n – 1 which satisfies

P1n()xi = yi , i = 1} n

Let P 2 n + 1 x be a Lagrange’s polynomial with degree n – 1 which satisfies

P2 n + 1()xi = yi , i = 2} n + 1

Now the polynomial P1  n + 1()x formed from the data set xi yi , i = 1} n + 1 can be obtained from the formula

xx– 1P2n+1()x – xx– n+1P1n()x P1 n + 1()x = ------xn+1–x1

- This is trivially true since P 1  n + 1 ()x is a polynomial of degree n satisfying P 1  n + 1 ()x i = y i , i = 1 } n + 1

- This can be generalized to any interval ij :

xx–ij+Pii +1 } ij+–1+ xi–xP i+1 i+2 } ij+ P ii +1 } ij+ = ------xi–xij+

Scientific computing III 2011: 6. Interpolation 16 Interpolation: polynomials

- Graphically: i i + 1 Pii +1 } ij+–1½ i + 2 ¾ŸPii +2 } ij+ } P ¿ i+1 i+2 } ij+ ij+–2 ij+–1 ij+

- For practical calculations it is better to simplify the notation:

Sij =Pii +1 i+2 } ij+ - Now the recursion relation becomes

xx–ij+Sij –1 + xi–xS i+1 j–1 Sij = ------with S i 0 = y i . xi–xij+ - For example

y 1 = S 10

S11

y2 = S20 S12

S21 S13

y3 = S30 S22

S31

y4 = S40

Scientific computing III 2011: 6. Interpolation 17

Interpolation: polynomials

- In Fortran90 this looks like program neville implicit none integer,parameter :: rk=selected_real_kind(15,100) real(rk),allocatable :: xa(:),ya(:),s(:,:) real(rk) :: x,y,x1,x2,dx integer :: n,i,j,ix,ixmax character(len=80) :: argu if (iargc()/=3) then call getarg(0,argu) write(0,’(a,a,a)’) ’usage: ’,trim(argu),’ x1 x2 dx’ stop end if call getarg(1,argu); read(argu,*) x1 ! First x call getarg(2,argu); read(argu,*) x2 ! Last x call getarg(3,argu); read(argu,*) dx ! Step of x read(5,*) n ! allocate(xa(n),ya(n),s(n,0:n-1)) ! Read in do i=1,n ! data xi yi read(5,*) xa(i),ya(i) ! end do ! s(1:n,0)=ya ixmax=(x2-x1)/dx do ix=0,ixmax ! Loop over x x=x1+dx*ix do j=1,n-1 ! Neville’s recursion loops to calculate Sij do i=1,n-j s(i,j)=((x-xa(i+j))*s(i,j-1)+(xa(i)-x)*s(i+1,j-1))/(xa(i)-xa(i+j)) end do end do write(6,*) x,s(1,n-1) end do stop end program neville

Scientific computing III 2011: 6. Interpolation 18 Interpolation: polynomials

- Example: sin x 4 points 8 points

Scientific computing III 2011: 6. Interpolation 19

Interpolation: polynomials

1 - Another example - -- x

4points 8 points 16 points

Scientific computing III 2011: 6. Interpolation 20 Interpolation: polynomials

- In practice the polynomial is often calculated not using the previous recursion relation but using the differences C mi

and Dmi :

Cmi =Pi} im+ –Pi} im+ –1, Dmi =Pi} im+ –P i+1 } im+ - Based on the recursion relation for the P ‘s it is easy to derive the recursion relation for these differences

xim++1–x Cmi +1–Dmi Dm+1i= ------xi–xim++1

x i – x C mi +1–Dmi C m + 1  i = ------xi–xim++1

- On level m the corrections D and C take the polynomial one degree higher.

- Final polynomial P 12}N is thus obtained by traversing the array starting from any y i and summing up the corrections.

x1 y1 = P1

P12

x2 y2 = P2 P123 Cm + 1 i P23 P1234 Dm + 1 i x3 y3 = P3 P234

P34

x4 y4 = P4

Scientific computing III 2011: 6. Interpolation 21

Interpolation: polynomials

- An example implementation: void polint(double xa[], double ya[], int n, double x, double *y, double *dy) { int i,m,ns=1; double den,dif,dift,ho,hp,w; double *c,*d; dif=fabs(x-xa[1]); c=vector(1,n); d=vector(1,n); for (i=1;i<=n;i++) { if ( (dift=fabs(x-xa[i])) < dif) { // Find the table entry nearest to x ns=i; dif=dift; } c[i]=ya[i]; d[i]=ya[i]; } *y=ya[ns--]; for (m=1;m

Scientific computing III 2011: 6. Interpolation 22 Interpolation: polynomials

• Neville’s algorithm does not give us the coefficients of the polynomial. - They can be calculated by using so called divided differences.

- The proof of the theorem on existence of interpolating polynomial provides us a method for constructing it: Newton’s algorithm

- We start at zero degree polynomial:

P0 x = y0 - Add the second term: Note: Now indexing of the data points starts from 0: xi yi , i = 012} N . P1 x = P0 x + a1 xx– 0 = y0 – a1 xx– 0 ;

y1–y0 P1 x1 =y1Ÿa1= ------x1–x0 - Add the third term:

P 2 x = P 1 x + a 2 xx– 0 xx– 1 = y 0 + a 1 xx– 0 + a 2 xx– 0 xx– 1

By setting P 2 x 2 = y 2 we obtain constant a 2 .

- Iteration for the k th polynomial is

P k x = P k – 1 x + a k xx– 0 xx– 1 } xx– k – 1

- The final polynomial going through all the points is

P N x = a 0 + a 1 xx– 0 +a 2 xx– 0 xx– 1 } ++a N xx– 0 xx– 1 } xx– N – 1 , where a 0 = y 0 .

Scientific computing III 2011: 6. Interpolation 23

Interpolation: polynomials

- Example:

Scientific computing III 2011: 6. Interpolation 24 Interpolation: polynomials

- The same data using the Lagrange polynomial:

Scientific computing III 2011: 6. Interpolation 25

Interpolation: polynomials

- Due to the uniqueness of the interpolating polynomial the Lagrange’s and Newton’s polynomials are the same: 1 1 1 1 Lagrange: Px =–36§·x – - -- x–1–16§·x – - -- x–1+14§·x – - -- §·x– - -- ©¹4 ©¹3 ©¹3 ©¹4 1 1 1 Newton: Px =2+ 36§·x – - -- –38§·x – - -- §·x– - -- ©¹3 ©¹3 ©¹4

Emacs package imaxima: uses TeX to output maxima results.

Scientific computing III 2011: 6. Interpolation 26 Interpolation: polynomials

- An efficient way to evaluate Newton’s polynomial is nested multiplication.

- Newton’s polynomial Px = a0 + a1 xx– 0 +a2 xx– 0 xx– 1 } ++aN xx– 0 } xx– N can be written in the form

Ni–1 Px =a+a xx– 0 ¦i– j i=1j=0

- Using the common terms xx–i for factoring we obtain the nested form

Px =a0+ xx–0 a1+ xx–1 a2+ xx–2 a3+} xx–N–1aN })

- For evaluating Px for xt = we begin from the innermost parenthesis

Q0=aN

Q1=Q0 tx–N–1+aN–1

Q2=Q1 tx–N–2+aN–2 }

QN=QN–1 tx–0+a0

- Q N is the value of the polynomial.

Scientific computing III 2011: 6. Interpolation 27

Interpolation: polynomials

- An efficient way to compute the coefficients of the Newton’s polynomial is based on divided differences.

- We have N + 1 points xi  fx i i = 0} N

- Newton’s polynomial interpolating these points is

Ni–1 P x =a xx– ( –1 xx– is interpreted as 1 .) N ¦i– j –j=0 j i=0j=0

- Coefficients a i do not depend on N: P N is obtained from P N–1 by adding one more term without changing PN–1 .

- A systematic way to obtain a i is to set x equal to points x i one at a time:

­fx 0 =a0 ° °fx 1 =a0+a1 x1–x0 ® °fx =a+a x–x+a x–x x–x °2 0 11 0 22021 ¯}

or in compact form ki–1 fx k=¦ai– xk–xj 0 kN i=0j=0

Scientific computing III 2011: 6. Interpolation 28 Interpolation: polynomials

- We can now see that a k depends on the values of f at points x0x1 } xk ; formally:

ak = fx>@0x1 } xk

- Quantity fx>@ 0x1 } xk is called the divided difference of order k for f .

Scientific computing III 2011: 6. Interpolation 29

Interpolation: polynomials

- Newton’s polynomial now takes the form

Ni–1 ­½ PN x =®¾fx>@0x1}xi xx–j ¦¯¿– i=0j=0

- Divided differences can be computed recursively:

ki–1 k–1 k–1 i–1 fx k=¦ai– xk–xj =fx>@0}xk – xk–xj+¦fx>@0}xi – xk–xj i=0j=0 j=0 i=0 j=0

i–1 fx –k–1fx>@}x x–x k ¦i=00 i –kj j=0 o fx>@0}xk= ------k–1 x–x –j=0kj

- Possible algorithm:

1. Set fx>@0 = fx 0

2. For k = 12} N compute fx>@0 } x k from the equation above.

- Well, there are more efficient ways to do this.

Scientific computing III 2011: 6. Interpolation 30 Interpolation: polynomials

- Recursive property of divided differences:

fx>@1}xk –fx>@0}xk–1 fx>@0}xk = ------xk–x0

- Uniqueness of the polynomial: no change if points x i  y i permuted

o The divided difference fx>@ 0 } x k is invariant under permutations of arguments x 0 } x k .

- This means that the recursive relation can be written as

fx>@i+1}xj–fx>@i}xj–1 fx>@ixi+1}xj–1xj = ------xj–xi

- Now the divided differences can be evaluated as follows:

fx>@i = fx i fx>@i + 1 = fx i + 1 fx>@i + 2 = fx i + 2

fx>@i+1–fx>@i fx>@i+2–fx>@i+1 fx>@ixi+1 = ------fx>@i+1xi+2 = ------xi+1–xi xi+2–xi+1

fx>@i+1xi+2–fx>@ixi+1 fx>@ixi+1xi+2 = ------xi+2–xi

Scientific computing III 2011: 6. Interpolation 31

Interpolation: polynomials

- We can now construct a similar table for f>@} ‘s as in the Neville’s method:

x0 fx>@0

fx>@0 x1

x1 fx>@1 fx>@0x1 x2

fx>@1 x2 fx>@0x1 x2 x3

x2 fx>@2 fx>@1x2 x3

fx>@2 x3

x3 fx>@3

- To summarize the story so far: Interpolating polynomial Ni–1 PN x =¦ai– xx–j =a0+a1 xx–0+a2 xx–0 xx–1 } ++aN xx–0 } xx–N–1 i=0j=0

where

ak=fx>@0}xk,

fx>@1}xk–fx>@0}xk–1 fx>@0}xk = ------xk–x0

fx>@k = fx k .

Scientific computing III 2011: 6. Interpolation 32 Interpolation: polynomials

- Example:

Scientific computing III 2011: 6. Interpolation 33

Interpolation: polynomials

Scientific computing III 2011: 6. Interpolation 34 Interpolation: polynomials

Scientific computing III 2011: 6. Interpolation 35

Interpolation: polynomials

- Using the bold face values from the final table we can construct the Newton’s polynomial

3i–1 ­½ PN x =®¾fx>@0x1}xi xx–j ¦¯¿– i=0j=0

=fx>@0 +fx>@0x1 xx–0++fx>@0x1x2 xx–0 xx–1 fx>@0x1x2x3 xx–0 xx–1 xx–2

11 3 3 1 ==§·3+- -- x – 1 +- -- x – 1 §·x– - -- –2 x–1§·x– - -- x–0 –---6 x3– 16x2 + 10x – 9 ©¹23 ©¹2 ©¹2 3

Scientific computing III 2011: 6. Interpolation 36 Interpolation: polynomials

- Algorithm that computes all the divided differences can now be constructed.

- There is no need to store all f>@ } because only fx>@0 , fx>@0 x1 , ..., fx>@ 0} xN are needed.

- Denote ai = fx>@0x1 } xi Newton’s polynomial can be written as

Ni–1 PN x =¦ai– xx–j i=0j=0

- We can use 1D array ai(0:N) to store divided differences - At each step the new divided difference is obtained by

ai–ai–1 aim ------xi–xij–

Scientific computing III 2011: 6. Interpolation 37

Interpolation: polynomials

- And in C:

void Coeffs(int n, double *x, double *y, double *a) { int i, j; A simpler but more memory consuming way:

for (i=0;i<=n;i++) a[i]=y[i]; double x[n+1],y[n+1],a[n+1][n+1]; for (i=0;i<=n;i++) a[i]=y[i]; for (j=1;j<=n;j++) for(j=1;j<=n;j++) for (i=n;i>=j;i--) for(i=0;i<=n-j;i++) a[i]=(a[i]-a[i-1])/(x[i]-x[i-j]); a[i][j]=(a[i+1][j+1]-a[i][j-1])/ (x[i+j]-x[i]); }

- Evaluation of the polynomial: a[0] double Evaluate(int n, double *x, 0 double *a, double t) { int i; a[1] double pt; pt = a[n]; 1 01 for (i=n-1;i>=0;i--) pt=pt*(t-x[i])+a[i]; a[2] return(pt); } 2 12 012

a[3] 3 23 123 0123

Scientific computing III 2011: 6. Interpolation 38 Interpolation: polynomials

- Example: - sin x in >@0 1.6875 - Interpolation based on 10 equidistant points

Polynomial coefficients

Scientific computing III 2011: 6. Interpolation 39

Interpolation: polynomials

- Interpolation error:

Scientific computing III 2011: 6. Interpolation 40 Interpolation: polynomials

• Inverse interpolation:

- Approximation to inverse of a function

- Values yi = fx i calculated for x0x1 } xN

- Form the interpolation polynomial

Ni–1 PN y =¦ci– yy–j i=0j=0

- It is easy to use the abovementioned routines to compute ci

- Inverse interpolation can also be used to locate the root of a given function.

Scientific computing III 2011: 6. Interpolation 41

Interpolation: polynomials

• Finally a word of warning on using high-degree polynomials: they do go through the data points but may oscillate wildly between them:

Scientific computing III 2011: 6. Interpolation 42 Interpolation: rational functions

• In many cases it is impossible to get a good interpolation using only polynomials - This means that the polynomial oscillates wildly between points - This may be caused by a singularity in the function to be interpo- lated. 1 - It need not be in the interpolating interval; it may be near it or y = ------1 + x2 not at all on the real axis. - In these cases a good choice might be a rational function

- We define the rational function R ii +1 } im+ so that it goes

through points xiyi } xim+yim+ :

P()x p + p x ++} p x P R ()x= ------P = ------0 1 P ii +1 } im+ Q()x Q Q q0+q1x++}qQx

- You have to give the order of the numerator and denominator when specifying the interpolating function. - Since there are PQ + + 1 unknowns we must have m + 1 = PQ+ + 1 - It is possible to derive a recursion relation as in the case of Neville’s algorithm:

R i+1 } im+ –Ri} im+ –1 R ii +1 } im+ = R i+1 } im+ + ------§·xx–i §·R i+1 } im+ –Ri} im+ –1 ¨¸------¨¸1– ------–1 ©¹xx–im+ ©¹R i+1 } im+ –R i+1 } im+ –1

- It is started with R i = y i and R ii +1 } im+ =0 , when m=–1 .

Scientific computing III 2011: 6. Interpolation 43

Interpolation: polynomials

• A sidenote: As expected, polynomial interpolation is used in computing transcendental functions in math libraries - E.g. function sin x from the libm sources:

/* @(#)k_sin.c 1.3 95/01/18 */ * For better accuracy, let /* * 3 2 2 2 2 * ======* r = x *(S2+x *(S3+x *(S4+x *(S5+x *S6)))) * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. * then 3 2 * * sin(x) = x + (S1*x + (x *(r-y/2)+y)) * Developed at SunSoft, a Sun Microsystems, Inc. business. */ * Permission to use, copy, modify, and distribute this * software is freely granted, provided that this notice * is preserved. * ======*/

/* __kernel_sin( x, y, iy) * kernel sin function on [-pi/4, pi/4], pi/4 ~ 0.7854 * Input x is assumed to be bounded by ~pi/4 in magnitude. * Input y is the tail of x. * Input iy indicates whether y is 0. (if iy=0, y assume to be 0). * * Algorithm * 1. Since sin(-x) = -sin(x), we need only to consider positive x. * 2. if x < 2^-27 (hx<0x3e400000 0), return x with inexact if x!=0. * 3. sin(x) is approximated by a polynomial of degree 13 on * [0,pi/4] * 3 13 * sin(x) ~ x + S1*x + ... + S6*x * where * * |sin(x) 2 4 6 8 10 12 | -58 * |------(1+S1*x +S2*x +S3*x +S4*x +S5*x +S6*x ) | <= 2 * | x | * * 4. sin(x+y) = sin(x) + sin’(x’)*y * ~ sin(x) + (1-x*x/2)*y

Scientific computing III 2011: 6. Interpolation 44 Interpolation: rational functions

- Below two examples where rational function interpolation works better than polynomial:

Scientific computing III 2011: 6. Interpolation 45

Interpolation: numerical differentiation

• Numerical differentiation is not a trivial task.

- Subtracting almost equal numbers causes loss of significant figures.

• The crudest approximation is based on the definition of the derivative fc :

fxh + –fx fc x | ------h

- Error truncation estimation can be obtained from the remainder term

fxh+ –fx 1 f ' x =------–- -- hf 2 [ h 2

- This converges as Oh ; for Oh 2 we have to take more terms.

- Consider the following Taylor’s series 1 1 1 ­fxh + =fx +hf' x ++++----- h 2 f 2 x ----- h 3 f 3 x ----- h 4 f 4 }x ° 2! 3! 4! ® 1 1 1 °fxh – =fx –hf' x +----- h 2 f 2 x–----- h 3 f 3 x++----- h 4 f 4 }x ¯ 2! 3! 4! - By subtraction we get 2 2 fxh + –fxh – =2hf' x +----- h 3 f 3 x+----- h 5 f 5 x 3! 5! fxh + –fxh – 1 1 o f' x =------–----- h 2 f 3 x–----- h 4 f 5 x–} i.e. Oh 2 ! 2h 3! 5!

Scientific computing III 2011: 6. Interpolation 46 Interpolation: numerical differentiation

- Including the truncation term we get

fxh+ –fxh– 1 f' x =------–- -- h 2 f 3 [ , [>@xh–xh+ 2h 6

- We can write

fxh+ –fxh– 1 1 f' x =------–----- h 2 f 3 x–----- h 4 f 5 x–} 2h 3! 5!

fxh()+ –fxh()– as f'()x =------+ a h 2 +++a h 4 a h 6 } 2h 2 4 6

- Here the coefficients a 2n only depend on f and x not on h. - A method called Richardson extrapolation is based on this fact.

- Let’s define function I in a following way ( f and x now constants)

fxh()+ –fxh()– I()h= ------2h

- This is the approximation to derivative f c x and its error behaves as Oh 2 . - Our goal is to compute the limit

lim I()h=f'()x ho0

Scientific computing III 2011: 6. Interpolation 47

Interpolation: numerical differentiation

- We can calculate I for any set of h n and use these as the approximation to fc x .

- However, we are smart and want to utilize the previous results in the series hn to calculate the next member.

- Assume we have approximations I h and I h e 2 . fxh()+ –fxh()– - As we have f'()x = ------+ a h 2 +++a h 4 a h 6 } we can write 2h 2 4 6 ­I()h= f'()x–ah2–ah4–ah6– } (1) ° 2 4 6 ® h h2 h4 h6 °I()- -- = f'()x–a§·- -- –a§·- -- –a§·- -- – } (2) ¯ 2 2©¹2 4©¹2 6©¹2 - Now calculate (1) – 4u (2) h 3 15 o I()4h – I()- -- = –3f'()x– - -- a h 4 – ------a h 6 – } 2 4 4 16 6 h 1 h 1 5 o I ()- -- +- -- I ()- -- –I()h =f'()x+ - -- a h 4 ++------a h 6 } 2 3 2 4 4 16 6

- So, by adding the term 13 e I>@()he2–I()h to I() he 2 we have obtained a more accurate estimation to fc x!

4 - We could go on by killing the error term a 4he4 from the series above. - This is called Richardson extrapolation (RE). (Extrapolation in the sense ho0 .)

Scientific computing III 2011: 6. Interpolation 48 Interpolation: numerical differentiation

- Generally: Assume we have a function I h : f I()h =La– h2k (1) ¦2k k=1

- Coefficients a 2k are unknown.

- Goal is to approximate quantity L using function I h : we want to calculate the limit

L=lim I h ho0

- Choose a value for h and calculate the numbers

h Dn I1 =§·----- , n=12} ©¹2n - According to (1)

h2k Dn 1 =L+ Ak 1§·----- , ¦ ©¹2n

where Ak 1=–a2k

- Dn 1 give a rough estimate for L.

Scientific computing III 2011: 6. Interpolation 49

Interpolation: numerical differentiation

- We can get more accurate estimates by RE using the recursion relation:

4m 1 Dnm() +1 =------Dnm() –------Dn()–1m 4m–1 4m–1

- It can be shown that the quantities Dnm  have the form

f h2k Dnm() =L+ Akm() §·----- ¦ ©¹2n km=

- Essential here is that the sum over k starts with term km= , so that the first error term is Oh 2m .

- Numbers Dnm  can be arranged into triangular form:

D()11 D()21 D()22 D()31 D()32 D()33 }}} DN()1 DN()2 DN()3 } DNN()

Scientific computing III 2011: 6. Interpolation 50 Interpolation: numerical differentiation

- A practical algorithm for RE is the following:

1. Write a subroutine for function I h . 2. Choose suitable values for N and h . 3. For i = 12} N compute Di I1 = he2i 4. For 1ddd ijN compute Dij() +1 =Dij()4 + j–1–1>@Dij() –Di()–1j

- The algorithm is easy to implement:

void Derivative(double (*f)(double x), double x, int N, double h, double D[NMAX][NMAX]) { int i, j; double hh; hh=h; for (i=0; i<=N; i++) { D[i][0]=(f(x+hh)-f(x-hh))/(2.0*hh); for (j=0; j<=i-1; j++) D[i][j+1] = D[i][j]+(D[i][j]-D[i-1][j])/(pow(4.0,j+1.0)-1.0); hh = hh/2.0; } }

Scientific computing III 2011: 6. Interpolation 51

Interpolation: numerical differentiation

- And the main program:

#include #include #include #define NMAX 10 double f(double x); void Derivative(double (*f)(double x), double x, int N, double h, double D[NMAX][NMAX]);

main (int argc, char **argv) { double x,h,D[NMAX][NMAX]; int N,i,j; if (argc!=4) { fprintf(stderr,"Usage: %s N x h\n",argv[0]); return (1); } N=atoi(*++argv); x=atof(*++argv); Function fx : h=atof(*++argv); if (N>=NMAX) { double f(double x) { fprintf(stderr,"N should be smaller than %d.\n",NMAX); return exp(-x*x); return(-1); } } for (i=0;i

Scientific computing III 2011: 6. Interpolation 52 Interpolation: numerical differentiation

- Results from the run: progs> ./richardson 6 1.0 0.5 i: 0 h/ 1 : -0.6734015585 i: 1 h/ 2 : -0.7203428752 -0.7359899807 i: 2 h/ 4 : -0.7319209458 -0.7357803026 -0.7357663241 i: 3 h/ 8 : -0.7348004908 -0.7357603391 -0.7357590082 -0.7357588921 i: 4 h/ 16 : -0.7355193541 -0.7357589753 -0.7357588843 -0.7357588824 -0.7357588823 i: 5 h/ 32 : -0.7356990047 -0.7357588882 -0.7357588824 -0.7357588823 -0.7357588823 -0.7357588823

2 2 2 - Accurate value: fx =e–x o fc x = –2xe–x o fc 1 = – - -- = –0.73575888 e

Scientific computing III 2011: 6. Interpolation 53

Interpolation: numerical differentiation

• Numerical derivatives can also be computed using the derivative if the interpolated polynomial:

fc x | Pc x

- In practice the interpolating polynomial is determined only using a couple of points near the argument. - For two nodes:

P1 x = fx 0 + fx>@0 x1 xx– 0

fx 1–fx 0 o fc x |Pc x = fx>@0x1 = ------x1–x0

- If x 0 = x and x 1 = xh+ we get

fxh + –fx f c x | ------h

- For x 0 = xh – and x 1 = xh+

fxh+ –fxh– f c x | ------(central difference form) 2h

Scientific computing III 2011: 6. Interpolation 54 Interpolation: numerical differentiation

- Now consider three nodes:

P2 x = fx 0 + fx>@0 x1 xx– 0 + fx>@0x1 x2 xx– 0 xx– 1

- The derivative is

P2c x = fx>@0 x1 + fx>@0x1 x2 2xx– 0 – x1

- This is the same as above but with the correction fx>@0x1 x2 2xx– 0 – x1 .

- The accuracy of the leading term fx>@ 0 x1 depends on the choice of x 0 and x1 .

- If xx= 0 + x1 e 2 o 2xx – 0 – x1 = 0 and correction term is zero. - This is the reason for the greater accuracy of the central difference form of derivative.

- In general we can analyze the error by using the error bound for the interpolating polynomial:

N f N+1 [ x fx –Px =------xx– N+1!– i i=0

Scientific computing III 2011: 6. Interpolation 55

Interpolation: numerical differentiation

- Let’s assume that P N is the polynomial of the least degree that interpolates f at points x0x1 x2 } xN .

- Now we can write the error bound as:

f N+1 [ fx –P x =------Z x N N+1! N where Z x =– xx–i and [[= x i=0

- Differentiation gives

1 d 1 fc x–Pc x =------Z x f N+1 [+------f N + 1 [ Zc x N N+1!dx N+1!

- If we can set x equal to a node the expression is simplified because Z xi =0

1 fc x =Pc x+------f N + 1 [ Zc x i N i N+1! i

- Or if we can choose x such that Zc x=0 :

1 d fc x=Pc x+------Z x f N+1 [ N N+1!dx

Scientific computing III 2011: 6. Interpolation 56 Interpolation: numerical differentiation

- Example: Derive the first-derivative formula using the interpolating polynomial P 3 x obtained with four nodes

x0x1 x2 x3 . Use central difference in choosing the nodes.

- Newton form of the polynomial

P3 x =fx 0 +fx>@0 x1 xx– 0 +fx>@0x1 x2 xx– 0 xx– 1 +fx>@0x1x2x3 xx–0 xx–1 xx–2

- And its derivative Note that: Pc x=fx>@x +fx>@xx^` xx– + xx– 3 01 012 0 1 Zc x = xx– +fx>@x x x ^` xx– xx– + xx– xx– + xx– xx– ¦– j 0 1 2 3 1 2 0 2 0 1 i jiz

- Formulas for the divided differences are obtained in a familiar way:

fx>@i=fx i

fx>@i+1–fx>@i fx>@ixi+1 = ------xi+1–xi

fx>@i+1xi+2–fx>@ixi+1 fx>@ixi+1xi+2 = ------xi+2–xi

Scientific computing III 2011: 6. Interpolation 57

Interpolation: numerical differentiation

- Now we choose the nodes using the central difference; i.e. x is in the middle of the nodes:

x0 = xh– , x1 = xh+ , x2 = x – 2h , x3 = x + 2h

- This allows us to write down the divided differences:

fx 1–fx 0fxh + –fxh – fx>@0x1 = ------= ------x1–x0 2h

fx 2–fx 1fx –2h –fxh + fx>@1x2 = ------= ------x2–x1 –3h

fx 3 –fx 2 fx +2h –fx –2h fx>@2x3 = ------= ------x3–x2 4h

fx>@1x2 –fx>@0x1 1 ­½fx –2h –fxh + fxh + –fxh – fx>@0x1x2 =------=------®¾------– ------x2–x0 – h ¯¿–3h 2h 1 =------>@ 2 fx –2h –3fxh – +fxh + 6h2

fx>@2x3 –fx>@1x2 1 ­½fx +2h –fx –2h fx –2h –fxh + fx>@1x2x3 =------=- -- ®¾------– ------x3–x1 h ¯¿4h –3h 1 =------>@ 3 fx +2h –4fxh + +fx –2h 12h2

Scientific computing III 2011: 6. Interpolation 58 Interpolation: numerical differentiation

- And the last one:

fx>@1x2x3 –fx>@0x1x2 fx>@0x1x2x3 =------x3–x0 1 ­½3 fx +2h –4fxh + +fx –2h 2 fx –2h –3fxh – +fxh + =------®¾------– ------3 h ¯¿12h2 6h2 1 =------>@fx +2h –2fxh + +2fxh – –fx –2h 12h3

- The following two factors are still needed:

xx–0+ xx–1=2xx–1–x0=0 2 xx–1 xx–2+ xx–0 xx–2+ xx–0 xx–1=–h

- The derivative of the polynomial

P3c x= fx>@0x1 +fx>@0x1x2^` xx–0+ xx–1

+fx>@0x1x2x3^` xx–1 xx–2+ xx–0 xx–2+ xx–0 xx–1

= fx>@0x1 +fx>@0x1x2x3^` xx–1 xx–2+ xx–0 xx–2+ xx–0 xx–1 fxh+ –fxh– fx+2h –2fxh+ +2fxh– –fx–2h = ------– >@------2h 12h

Scientific computing III 2011: 6. Interpolation 59

Interpolation: numerical differentiation

- Approximation for the derivative is then

fx–2h –8fxh– +8fxh+ –fx+2h fc x | ------12h

- This can be used in the Richardson extrapolation to speed up the convergence.

- Example: fx = sinx , x = 1.2309594 , h = 1

D(n,0) D(n,1) D(n,2) D(n,3) D(n,4) 0.32347058 0.33265926 0.33572215 0.33329025 0.33350059 0.33335248 0.33333063 0.33334409 0.33333365 0.33333335 0.33333317 0.33333401 0.33333334 0.33333334 0.33333334

- For comparison when only the first term fc x | >@fxh + – fxh – e 2h is used we get the result

D(n,0) D(n,1) D(n,2) D(n,3) D(n,4) 0.28049033 0.31961703 0.33265926 0.32987195 0.33329025 0.33333232 0.33246596 0.33333063 0.33333332 0.33333333 0.33311636 0.33333317 0.33333333 0.33333334 0.33333334

Scientific computing III 2011: 6. Interpolation 60 Interpolation: Hermite interpolation

• The interpolating function goes through data points xi fx i , i = 01} N :

Px i = fx i

• We can generalize this by adding requirements that the interpolating function also 2 reproduces derivatives: f1 x =x

mi mi P xi =f xi .

- The polynomial P is osculating to the function f .

- We have N+ 1points xi yi and nonnegative integers m0m1 }mN .

- The polynomial fulfilling these conditions has a degree at most

N Mm=i+N. ¦ 2 4 i=0 f2 x =x –x

k k - So our osculating polynomial interpolates the function f such that f1 0 =f2 0 for k =012 k k d d Px i = fx i for each i=01 }N and k=01}mi . dxk dxk

Scientific computing III 2011: 6. Interpolation 61

Interpolation: Hermite interpolation

- Note:

1) When N = 0 we have the m0 th Taylor polynomial for f at x0 .

2) When m i = 0 for i = 0} N we have the N th order interpolating polynomial for f

- When m i = 1 for i = 0} N the interpolating polynomials are called Hermite polynomials:

Px i = fx i and Pc xi = fc xi .

- One can derive the following form for the polynomial using the divided differences1 2 2 2 2 HN x =Q00 +Q11 xx–0+++Q11 xx–0 Q33 xx–0 xx–1 Q44 xx–0 xx–1 2 2 2 2 ++}Q2n+12 n+1 xx–0 xx–1 xx–2} xx–n–1 xx–n

- Coefficients Q ii are obtained by the following algorithm:

1. For i= 01}N do

a) Set z2i=xi, z2i+1=xi , Q2i0=fx i , Q2i+01=fx i , Q2i+11=fc xi

Q2i0–Q2i–10 b) If i z0 set Q2i1= ------z2i–z2i–1

Qij –1–Qi–1j–1 2. For i = 23} 2 N + 1 , for j = 23 } i set Q ij = ------zi–zij–

1. See e.g. Burden & Faires: Numerical Analysis, Ch. 3.5

Scientific computing III 2011: 6. Interpolation 62 Interpolation: Hermite interpolation

- An in Fortran90 it reads program hermite end do implicit none do i=2,2*n+1 integer,parameter :: & do j=2,i &rk=selected_real_kind(15,100) q(i,j)=(q(i,j-1)-q(i-1,j-1))/(z(i)-z(i-j)) end do real(rk),allocatable :: xa(:),ya(:),& end do &da(:),q(:,:),z(:) real(rk) :: x,y,x1,x2,dx,p ! Use the coefficients to calculate P(x) integer :: n,i,j,ix,ixmax,jmax,iargc ixmax=(x2-x1)/dx character(len=80) :: argu do ix=0,ixmax call getarg(1,argu); read(argu,*) x1 x=x1+dx*ix call getarg(2,argu); read(argu,*) x2 y=q(0,0)+q(1,1)*(x-xa(0)) call getarg(3,argu); read(argu,*) dx do i=2,2*n+1 read(5,*) n p=q(i,i) allocate(xa(0:n-1),ya(0:n-1),da(0:n-1),& jmax=i/2-1 &q(0:2*n-1,0:2*n-1),z(0:2*n-1)) if (ix==0) write(0,*) i,jmax do i=0,n-1 do j=0,jmax read(5,*) xa(i),ya(i),da(i) p=p*(x-xa(j))**2 end do end do n=n-1 if (mod(i,2)==1) p=p*(x-xa(jmax+1)) q=0.0 y=y+p z=0.0 end do do i=0,n write(6,’(f10.4,g20.10)’) x,y z(2*i)=xa(i) end do z(2*i+1)=xa(i) q(2*i,0)=ya(i) stop q(2*i+1,0)=ya(i) end program hermite q(2*i+1,1)=da(i) if (i/=0) then q(2*i,1)=(q(2*i,0)-q(2*i-1,0))/& &(z(2*i)-z(2*i-1)) end if

Scientific computing III 2011: 6. Interpolation 63

Interpolation: Hermite interpolation

- Graphically: interpolating fx =e–x2 :

- Hermite polynomial is often used in the form of cubic Hermite polynomial:

­Pa =fa ° °Pc a =fc a ® °Pb =fb ° ¯Pc b =fc b

- The divided difference formula for this is

2 H3 x =fa + xa– f' a– xa– faab>@ 2 + xa– xb– faabb>@

Note: The following definition applies to divided differ- ences with repetition of the same argument n : n–1 f x fx>@x}x= ------i iii n–1!

Scientific computing III 2011: 6. Interpolation 64 Interpolation: piecewise interpolation

• High degree polynomial o oscillations.

• Remedy: construct interpolating polynomials for subintervals of the data ^`xi fy i : piecewise polynomials or splines

- The simplest spline is of the first order: linear interpolation

­S0 x xt>@0t1 ° °S1 x xt>@1t2 - In explicit form: Sx =® ° } ° ¯SN–1 x xt>@N–1tN

Scientific computing III 2011: 6. Interpolation 65

Interpolation: piecewise interpolation

- Where each piece of S is linear:

Si x = aixb– i

- In general a spline S of degree k fullfills the following conditions

1. The domain of S is interval >@ab 2. S, Sc, Scc , S 3 , } , S k – 1 are all continuous on >@ab .

3. There are points xi (knots of S) such that ax= 0 x1 } xn = b and that S is

a polynomial of degree at most k on each subinterval >@xi xi + 1 .

- A quadratic spline Qx is a continuously differentiable piecewise quadratic function.

- Example:

­x2 xd0 ° Qx =®–x2 0ddx 1 ° ¯12–x xt1

Scientific computing III 2011: 6. Interpolation 66 Interpolation: piecewise interpolation

- Ensure that the function Qx really is a quadratic spline. Check all knots:

lim Qx = limx2 = 0 lim Qx = lim–x2 = 0 xo0– x o 0– xo0+ x o 0+ lim Qx = lim–x2 = – 1 limQx = lim 12–x= – 1 xo1– x o 1– xo1+ xo1+ lim2Qc x =lim0x= lim2Qc x = lim–x= 0 xo0– xo0–xo0+ xo0+ lim2Qc x=lim2–x=–lim2Qc x = lim –= – 2 xo1– xo1–xo1+ xo1+

- Let’s assume we have the data set ^`xi yi , i = 01} N and we want to interpolate it using quadratic splines. 2 o we have to determine N functions Qi x =aix +bixc+i , i=01}N–1 . o we have to determine 3 N coefficients

- On each subinterval >@x ix i+1Q must satisfy:

Qi xi=yi and Qi xi+1=yi+1 o 2N equations

- Continuity of the first derivative gives

Qci–1 xi=Qci xi, i=12}N–1 o N– 1equations

- Only 3N –1 equations. Additional conditions e.g. Qc x0=0 .

Scientific computing III 2011: 6. Interpolation 67

Interpolation: piecewise interpolation

- So, we seek for a piecewise quadratic function

­Q0 x xx>0x1 ° °Q1 x xx>1x2 Qx =® °} ° ¯QN–1 x xx>N–1xN@

which is continuously differentiable on the interval >@ x0xN and which interpolates the table: Qx i=yi

- Denote zi=Qc xi . We can write

z –z i+1 i 2 Qi x=------xx–i +zi xx–i+yi 2 xi+1–xi

- It easy to verify that

Qi xi=yi

Qci xi=zi

Qci xi+1=zi+1

Scientific computing III 2011: 6. Interpolation 68 Interpolation: piecewise interpolation

- In order to Q to be continuous and interpolate the data table ^` xi yi it is necessary and sufficient that

Qi xi + 1 = yi + 1

- This gives us

§·yi + 1 – yi zi + 1 =– zi +2¨¸------, 0 ddiN –1. ©¹xi + 1 – xi

T - The vector >@ z0z1}zN can now be constructed by starting from (an arbitrary) value of z0=Qc x0

- Well, the recursion relation above is a group of linear equations (though a trivial one):

z Qc x 1000}0 0 0 z y–ye x–x –1 1 00}0 1 1 0 10 z y–ye x–x 01– 10}0 3 = 2121 001–1}0 } } } zN–1 yN–1–yN–2e xN–1–xN–2 00}01–1 zN yN–yN–1e xN–xN–1

Scientific computing III 2011: 6. Interpolation 69

Interpolation: piecewise interpolation

- The program that does the interpolation: program qspline z(0)=Qp0 do i=0,n-1 implicit none z(i+1)=-z(i)+2.0*(ya(i+1)& integer,parameter :: rk=selected_real_kind(15,100) & -ya(i))/(xa(i+1)-xa(i)) end do real(rk),allocatable :: xa(:),ya(:),z(:) real(rk) :: x,y,x1,x2,dx,Qp0 ixmax=(x2-x1)/dx integer :: n,i,j,ix,ixmax character(len=80) :: argu do ix=0,ixmax x=x1+dx*ix if (iargc()/=4) then if (xxa(n)) cycle call getarg(0,argu) do i=0,n-1 write(0,’(a,a,a)’) & if (x>=xa(i).and.x

read(5,*) n allocate(xa(0:n),ya(0:n),z(0:n)) do i=0,n-1 read(5,*) xa(i),ya(i) end do n=n-1

Scientific computing III 2011: 6. Interpolation 70 Interpolation: piecewise interpolation

- Random sets of knots created with gawk one-liner: gawk ’BEGIN {srand(); for (x=0;x<=5.01;x+=0.5) print x,rand()}’ - Graphically:

11 knots 41 knots

Scientific computing III 2011: 6. Interpolation 71

Interpolation: piecewise interpolation

• Splines of degree 3 are cubic splines and they a probably the most commonly used splines.

- Cubic splines have more flexibility than quadratic splines o ‘smoother’ interpolant.

- In general for splines the following theorem applies:

m Let fx C >@ab and S 2m – 1 x be the interpolating spline of f. If gx is an m times continuously

differentiable function that fulfills the same interpolating conditions as S 2m–1 then

b b 2 m 2 S m x dxd g x dx ³ 2m–1 ³ a a

- For cubic splines (m=2 ) we can write this in the form

b b 2 2 S'' x dxd g'' x dx ³ 3 ³ a a

- Now the curvature of a curve yx is defined as

y'' x N x = ------232e ^`1+>@y' x

- If we assume that y ' x « 1 in >@ab  we can write N x | y '' x

Scientific computing III 2011: 6. Interpolation 72 Interpolation: piecewise interpolation

- So, for cubic splines we can say that they are the interpolants that have — on the average — the smallest curvature.

- Another way to depict this is to imagine the spline as a solid wire that is forced to go through the knots and minimizes its bending energy.

- As usual we have the data set or knots

x x0 x1 } xN

y y0 y1 } yN

- We assume that knots are distinct and in ascending order: x0 x1 } xN

- The interpolating function Sx consists of N cubic polynomials:

­S0 x xx>0x1 ° °S1 x xx>1x2 Sx =® i.e. S i x interpolates the subinterval >@xixi+1 °} ° ¯SN–1 x xx>N–1xN@

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Interpolation: piecewise interpolation

- The interpolation conditions are

Si xi = yi , 0 ddiN

- The continuity conditions imposed only at the interior knots x1 x2 } xN – 1 are

lim S k x = lim S k x , k = 012 – i + i xxoi xxoi

- Are these enough:

We have N 3rd order polynomials o 4N parameters. Interpolation conditions o N + 1 equations. Continuity conditions o 3 N – 1 equations. Total 4 N – 2 equations.

o need two more equations.

- A common choice is

S'' x0 = S'' xN = 0

- The resulting function is termed a natural cubic spline.

Scientific computing III 2011: 6. Interpolation 74 Interpolation: piecewise interpolation

- Example: Determine parameters a , b , c , d , e , f , g , and h so that Sx is a natural cubic spline, where

­ax3 +bx2 ++cxd x>@–10 Sx =® ¯ex3 +fx2 ++gxh x>@01

and with interpolation conditions: S –1=1 , S 0=2 , S 1=–1 .

- Let the two polynomials be S 0 x and S1 x . - From interpolation conditions we get

S0 0=d=2

S1 0=h=2

S0 –1 =–a+bc– =–1

S1 1=efg++3=– - The 1st derivative of Sx is ­3ax2 +2bxc+ S' x=® ¯3ex2 +2fxg+ - Continuity of S' x implies S'0 =cg= - The 2nd derivative is ­6ax + 2b S'' x = ® ¯ 6ex + 2f

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Interpolation: piecewise interpolation

- From the continuity of S '' x we get S''0 = bf= - And the two extra conditions ­S''1 – =–6a+2b=0 ® ¯S''1 =–6e+2f=0 - From all these equations we get a=–1, b = –3 , c = –1 , d = 2, e = 1, f = –3 , g = –1 , h = 2

Scientific computing III 2011: 6. Interpolation 76 Interpolation: piecewise interpolation

- In what follows we will develop a general algorithm to determine the natural cubic spline.

- Based on the continuity of S '' x we can unambiguously define the following numbers

zi { S'' xi , 0 ddiN

- The two extra conditions say that z 0 = zN = 0 but the other values are still unknown.

- We know that on each subinterval >@ xi xi + 1 the second derivative S '' x is linear, and takes the values z i and z i + 1 at the end points:

zi+1 zi Si'' x =------xx–i+---- x i + 1 – x , hi hi

where we have defined hi=xi+1–xi , 0ddiN–1

- Integrating this twice we get

zi+1 2zi 2 Si' x=------xx–i–------x i + 1 – x +c 2hi 2hi

zi+1 3zi 3 Si x=------xx–i +------x i + 1 – x ++cxd 6hi 6hi where c and d are integration constants.

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Interpolation: piecewise interpolation

- By adjusting the constants we get a more convenient form of the spline

z z i+1 3i 3 Si x =------xx–i +------x i + 1 – x ++Ci xx–i Di xi+1–x 6hi 6hi

- Constants C i and D i can be determined by applying the interpolation conditions:

Si xi=yi , Si xi+1=yi+1

z z yh z yh z i+1 3i 3§·i+1i i + 1 §·ii i o Si x=------xx–i +------x i + 1 – x ++¨¸------– ------xx–i ¨¸---- – ------xi+1–x 6hi 6hi ©¹hi6 ©¹hi6

- We still have to find out the values of zi . - This is done by applying the remaining condition – namely the continuity of S' x :

- At the interior knots xi , 1 ddiN–1 , we must have

S'i–1 xi=S'i xi

- From above we get the derivative:

z z yh z y h z i+1 2i 2i+1i i + 1 i i i S'i x= ------xx–i –------x i + 1 – x + ------– ------– ---- + ------2hi 2hi hi6 h i 6

Scientific computing III 2011: 6. Interpolation 78 Interpolation: piecewise interpolation

- And

hi h i y i + 1 – y i S'i xi = – ---- z i + 1 – ---- z i + b i , b i = ------6 3 h i

hi–1 h i – 1 y i – y i – 1 S ' i – 1 x i = – ------z i – 1 – ------z i + b i – 1 , b i – 1 = ------6 3 h i – 1

- When we set these equal we get – after a little rearrangement

h i – 1 z i – 1 + 2 h i – 1 – h i z i +6h i z i + 1 = b i – b i – 1 , 1 ddiN– 1

- By letting

­u i = 2 h i – 1 + h i ® ¯v i =6 b i – b i – 1

we get a tridiagonal system of equations for zi :

­z0=0 ° ®hi–1zi–1+uizi+hizi+1=vi1ddiN–1 ° ¯zN=0

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Interpolation: piecewise interpolation

- In matrix form this is

10 z0 0

h0u1h1 z1 v1 huh z v 122 2 = 2 } } }

hN–1uN–1hN–1zN–1 vN–1

0 1 zN 0

- We can eliminate the first and the last equations and get finally the form:

u1h1 z1 v1

h1u2h2 z2 v2 }} }=}

hN–3uN–2hN–2zN–2 vN–2

hN–2uN–1zN–1 vN–1

which is a symmetric tridiagonal system of order N–1 .

Scientific computing III 2011: 6. Interpolation 80 Interpolation: piecewise interpolation

- This system can be readily solved by

2 ­ h ° u m u – ------i – 1 ° i i u the forward elimination: ® i – 1 i =23}N–1 ° h v ° i – 1 i – 1 v i m v i – ------¯ u i – 1

­ vN–1 ° zN–1m ------° uN–1 and the back substitution: ® iN= –2N–3 }1 ° v i – h i z i + 1 ° z i m ------¯ u i

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Interpolation: piecewise interpolation

- Finally we can write down the algorithm for the natural cubic spline for data set ^`xi yi , i = 01} N :

1. For i = 01 } N – 1 compute

­ hi = xi + 1 – xi ° ® yi + 1 – yi ° bi = ------¯ hi 2. Set

­u 1 = 2 h 1 + h 0 ® ¯v 1 =6 b 1 – b 0 3. For i=23 }N–1compute 2 ­ hi–1 ° ui= 2 hi+hi–1– ------° ui–1 ® ° hi–1vi–1 ° v i = 6 b i – b i – 1 – ------¯ ui–1 4. Set

­ z N = 0 ® ¯ z 0 = 0 5. For iN= – 1 N – 2 } 1 compute

v i – h i z i + 1 z i = ------. u i

Scientific computing III 2011: 6. Interpolation 82 Interpolation: piecewise interpolation

- This algorithm might fail if u i = 0 for some i .

- It is clear that u1 !!h1 0 .

- Now if ui – 1 ! hi – 1 then ui ! hi because 2 hi–1 ui =2 hi + hi – 1 – ------!!2 h i + h i – 1 –hi–1hi ui–1

- So, by induction: ui!0 , i=12}N–1 .

- Now we have the coefficients z i of the spline

z z yh z yh z i+1 3i 3§·i+1i i + 1 §·ii i Si x=------xx–i +------x i + 1 – x ++¨¸------– ------xx–i ¨¸---- – ------xi+1–x 6hi 6hi ©¹hi6 ©¹hi6

- However, this is not the best possible form to evaluate the spline.

- In order to evaluate it using the nested multiplication we have to massage it into the form

2 3 Si x=Ai+Bi xx–i++Ci xx–i Di xx–i

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Interpolation: piecewise interpolation

- Notice that the equation above is in the form of Taylor expansion of S i around point xi , Hence,

1 1 A = S x , B = S ' x , C = - -- S '' x , D = - -- S ''' x i i i i i i i 2 i i i 6i i

- Therefore z A=y and C= ---i . i i i 2

3 - The coefficient of x in the form above is D i and in the previous form z i + 1 – z i e 6 h i

z i + 1 – z i o D i = ------6 h i

- Finally the value S i ' x i is give as

h i z i + 1 h i z i y i + 1 – y i B i = – ------– ------+ ------6 3 h i

- The nested form of S i x is then finally – after a few steps of algebra:

­½zi zi+1–zi S i x =yi+ xx–i®¾Bi+ xx–i--- + xx–i------¯¿2 6hi

Scientific computing III 2011: 6. Interpolation 84 Interpolation: piecewise interpolation

- Then to the implementation.

- The following routine solves the coefficients z i given the input parameters xi , yi , N .

void Spline3_Coeff(int n,double *x,double *y,double *z) { int i; double *h, *b, *u, *v; h = (double *) malloc((size_t) (n*sizeof(double))); b = (double *) malloc((size_t) (n*sizeof(double))); u = (double *) malloc((size_t) (n*sizeof(double))); v = (double *) malloc((size_t) (n*sizeof(double))); for (i=0; i0; i--){ z[i] = (v[i]-h[i]*z[i+1])/u[i]; } z[0]=0; free(h); free(b); free(u); free(v); }

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Interpolation: piecewise interpolation

- And the evaluation routine computes the spline value at x0 given as an input argument.

double Spline3_Eval(int n, double *x, double *y, double *z, double x0) { int i; double h, tmp, result; for (i=n-1; i>=0; i--) { if (x0-x[i]>=0) break; } h = x[i+1]-x[i]; tmp = 0.5*z[i] + (x0-x[i])*(z[i+1]-z[i])/(6.0*h); tmp = -(h/6.0)*(z[i+1]+2.0*z[i])+(y[i+1]-y[i])/h + (x0-x[i])*tmp; result = y[i] + (x0-x[i])*tmp; return(result); }

Scientific computing III 2011: 6. Interpolation 86 Interpolation: piecewise interpolation

- Example: Function sin x at the interval >@0 1.6875 , with ten equidistant knots:

Comparison with Newton polynomial

Scientific computing III 2011: 6. Interpolation 87

Interpolation: piecewise interpolation

- Another example: The serpentine curve

x y = ------14e +x2

- And in parametric form

­ 1 °x = - -- tanT ® 2 ° ¯y = sin2T

- Let’s take knots evenly spaced in T so that they are non-uniformly spaced in x axis:

S T = n------, n = – 5 } 5 12

- We can interpolate the parametric form by constructing splines for both x T and y T - Note that the knots must be in ascending order.

Scientific computing III 2011: 6. Interpolation 88 Interpolation: piecewise interpolation

- Below are the results comparing the interpolation with a spline and with a polynomial.

Scientific computing III 2011: 6. Interpolation 89

Interpolation: Bézier curves

• One way of writing the interpolating polynomial is by using basis functions bi x :

N Px = cb x ¦ ii i=0

- Note that if we have one polynomial going through all data points it is unique and and does not depend on the basis func- tions used.

- As examples of basis functions we have used are monomials

i–1 bi x =x , i=0}N

and Lagrange’s polynomials

N §·xx–j bi()x = ¨¸------, i=0}N –©¹x–x j=0i j jiz

- Note that here the monomials do not depend on the data set but Lagrange’s polynomials depend on the positions of the

nodes ^`xi.

Scientific computing III 2011: 6. Interpolation 90 Interpolation: Bézier curves

- Graphically

Monomials

Lagrange

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Interpolation: Bézier curves

• The Weierstrass approximation theorem:

For any continuous function f on a closed interval, say >@01 , and any H ! 0 , there is a polynomial P such that for any x  >@ 01 fx – Px H

n - It was proven by Bernstein using polynomials Bi x :

n i f§·- -- Bn xofx , when nof , x>@01 ¦©¹n i i=0

where the Bernstein polynomial (BP) is

n n n! Bn x =§· 1–xni–xi, §·= ------i ©¹i ©¹i i! ni–!

- These can be generalized to any interval >@ab :

ni– i n §·n bx– xa– B i x = ------©¹i ba– n

i - It is easy to show that the polynomial B n x has a maximum at the position xa=+- -- ba– i n

Scientific computing III 2011: 6. Interpolation 92 Interpolation: Bézier curves

- Below are examples of various BP’s at the interval >@01

5 Bi

4 Bi

3 3 3 2 3 2 B0 x = 1–x B1 x=3x 1–x B2 x=3x 1–x 3 B 3 3 i B3 x=x

2 B2 x = 1–x2 B2 x=2x 1–x B2 x=x2 Bi 0 1 2

1 1 1 B0 x =1–x B1 x =x Bi

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Interpolation: Bézier curves

- BP’s are often used as a basis for interpolation of function fx :

N P x = pBN x N ¦ i i i=0

where coefficients p i are obtained from interpolation conditions:

PN xi =fx i , i=01}N

- Why are BP’s useful?

Coefficients p i give a hint of how the interpolate P N looks like.

- Let’s associate with each p i a control point t ipi such that

i t =a+- -- ba– i n

- Note that t 0p 0= x0y0 and t NpN= xNyN but the control points between these limits do not necessarily

correspond to the data set ^`xiyi, i=1}N–1 .

Scientific computing III 2011: 6. Interpolation 94 Interpolation: Bézier curves

- In Fortran90 the interpolation is done as follows: program bernstein mx=1.0/dx implicit none integer,parameter :: & do m=0,mx rk=selected_real_kind(12,100) x=dx*m y=0.0 real(rk),allocatable :: b(:),p(:) call bernstein_poly(n,x,b) real(rk) :: x,y,dx do i=0,n integer :: iargc,n,mx,i,m,k write(i+10,’(2g18.10)’) x,b(i) character(len=80) :: argu,file1,file2 y=y+p(i)*b(i) end do if (iargc()<5) then write(200,’(2g18.10)’) x,y write(0,’(a,a)’) ’usage: bernstein ‘,& end do ‘cpointfile interpfile dx n p0 p2 p3 .. pn’ stop close(100) end if close(200)

call getarg(1,file1) stop call getarg(2,file2) open(100,file=file1,status=’new’) end program bernstein open(200,file=file2,status=’new’) call getarg(3,argu); read(argu,*) dx call getarg(4,argu); read(argu,*) n allocate(b(0:n),p(0:n)) do i=0,n call getarg(i+5,argu); read(argu,*) p(i) write(100,’(2g18.10)’) & real(i,rk)/real(n,rk),p(i) end do

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Interpolation: Bézier curves

- Program uses the polpak library for calculating Bernstein polynomials. - The library is available in Fortran90, C and Matlab and can be downloaded from

http://www.csit.fsu.edu/~burkardt/f_src/polpak/polpak.html

- Compiling and linking a program using the library is easy (after you have created the library itself and assuming the library is in the current directory):

gfortran -o bezier bezier.f90 -L. -lpolpak

Scientific computing III 2011: 6. Interpolation 96 Interpolation: Bézier curves

- Now plot the control points ^`ti pi , i = 0} N and the interpolating polynomial PN x

- For example:

- This means that by specifying the control points we can easily construct a smooth curve that resembles a line going through those points.

- The more we have points the nearer the curve comes to the control points.

- Note that here we are no more talking about interpo- lation. - The polynomial does not go through the control points.

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Interpolation: Bézier curves

- What if we have an arbitrary curve on the xy  plane (i.e. not a function y= fx )? y - We can express the curve in the parametric form

­x= xt ® , t  >@ 01 (for example) x ¯y= yt

- If the set of control points is ^` xiyi the interpolated curve is

­ N x °P t = xBN t ­ x ° N i i xP= t ¦ ° N ° i=0 ® where ® y N °yP= t ° y ¯ N °P t = yBN t ° N ¦ i i ¯ i=0

- This curve is called the Bézier curve

- Note that ^` x iyi are now the coefficients of the ‘interpolating’ polynomials.

Scientific computing III 2011: 6. Interpolation 98 Interpolation: Bézier curves

- The previous program only needs a little modification to compute the Bézier curve: program bezier mt=1.0/dt implicit none integer,parameter ::& do m=0,mt & rk=selected_real_kind(12,100) t=dt*m x=0.0 real(rk),allocatable :: b(:),xa(:),ya(:) y=0.0 real(rk) :: x,y,t,dt call bernstein_poly(n,t,b) integer :: iargc,n,mt,i,m,k do i=0,n character(len=80) :: argu,file1,file2 y=y+ya(i)*b(i) end do if (iargc()/=2) then call bernstein_poly(n,t,b) write(0,’(a)’) ’usage: bezier n dt’ do i=0,n stop x=x+xa(i)*b(i) end if end do write(6,’(2g18.10)’) x,y call getarg(1,argu); read(argu,*) n end do call getarg(2,argu); read(argu,*) dt n=n-1 stop allocate(b(0:n),xa(0:n),ya(0:n)) do i=0,n end program bezier read(5,*) xa(i),ya(i) end do

Scientific computing III 2011: 6. Interpolation 99

Interpolation: Bézier curves

- Two examples:

Scientific computing III 2011: 6. Interpolation 100 Interpolation: B-splines

• The basis functions b i x of the interpolate (note the slight change in notation: Nko ) k Px = ¦cibi x i=0 may also be splines itself: they are so called B-splines (B for basis).

k - A B-spline of order k is denoted as N i x and it depends on the set of nodes xi . 3 For example in the case of cubic B-splines the functions N i x have two continuous derivatives

and span four points around xi .

- B-splines can be computed recursively

1 xdxx 0 ­ i i+1 Ni x = ® ¯0 otherwise

xx– x –x k §·i k–1§·ik++1 k–1 Ni x =¨¸------Ni x+¨¸------Ni+1 x, where k =12} and i=01r r}2 ©¹xik+–xi ©¹xik++1–xi+1

- B-splines can be used in a similar fashion as Bézier curves. - With B-splines one can model continuous curves with points where derivatives are discontinuous.

- For more information see Kahaner, Moler, Nash: Numerical Methods and Software, or Haataja et al., Numeeriset menetelmät käytännössä.

Scientific computing III 2011: 6. Interpolation 101

Interpolation: 2D

• Let the function depend on two variables: y= fx 1 x2

i j - The interpolation data is now a set of points in a rectangular grid ^`yijx1 x2 , i=1}m , j=1}n

- Let’s use the following notation:

ya(j,k) = y()x1a(j) x2a(k)

- First we must find the right position in the grid where 4 3 (j,k+1) (j+1,k+1) x1a(j) ddx1 x1a(j+1) x  x x2a(k) ddx2 x2a(k+1) 1 2

- Also let the function values at the corners be 1 2 (j,k) (j+1,k) y1 { ya(j,k)

y2 { ya(j+1,k)

y3 { ya(j+1.k+1)

y4 { ya(j,k+1)

Scientific computing III 2011: 6. Interpolation 102 Interpolation: 2D

- The desired y value is obtained by using bilinear inter- polation:

x – x1a(j) t = ------1 x1a(j+1)– x1a(j) x – x2a(k) u = ------2 x2a(k+1)– x2a(k)

yx(1  x 2 )1= – t 1 – u y 1 + t 1 – u y 2 ++tuy3 1 – t uy4

- For higher accuracy/smoothess one can generalize the spline interpolation: 1 1. Construct m 1D interpolates 2 in x direction. } 2. Based on these construct one 1D interpolate at the desired y value. m – 1 - For more details see e.g. Numerical Recipes paragraph m 3.6 : desired interpolation point : spline in x direction : spline in y direction

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Interpolation: practical tools

• If you are using C then GSL has many interpolation routines: Spline interpolation #include #include #include #include #include

int main (int argc, char **argv) { int i,nk,np; double *x,*y,x1,x2,dx,xi,yi; gsl_interp_accel *acc; //Index look-up acceleration gsl_spline *spline; if (argc!=4) { BEGIN { fprintf(stderr,"Usage: %s x1 x2 dx\n",argv[0]); return (1); srand(); } x1=0; x1=atof(*++argv);a x2=100.0; x2=atof(*++argv); dx=1.0; dx=atof(*++argv); n=(x2-x1)/dx+1; fscanf(stdin,"%d",&nk); print n; x = (double *) malloc((size_t) (nk*sizeof(double))); for (i=0;i "interpdata.dat"; spline = gsl_spline_alloc(gsl_interp_cspline,nk); } gsl_spline_init(spline,x,y,nk); }

np=(x2-x1)/dx; for (i=0;i gcc gsl_spline.c -lgsl -lgslcblas xi=x1+dx*i; gsl> gawk -f makeinterpdata.awk |a.out -1 101 0.01 > interpdata.int yi=gsl_spline_eval(spline,xi,acc); gsl> xgraph -P interpdata.* printf("%g %g\n",xi,yi); } gsl_spline_free(spline); gsl_interp_accel_free(acc); return 0; }

Scientific computing III 2011: 6. Interpolation 104 Interpolation: practical tools

• For Fortran users the is e.g. the SLATEC package: http://www.netlib.org/slatec/

- At http://www.csit.fsu.edu/~burkardt/f_src/slatec/slatec.html you can find some documentation and test programs.

- Installed on punk.it.helsinki.fi at /usr/local/lib (linking: gfortran prog.f90 -lslatec)

- If you want to build the library yourself

1) Download the package: http://www.physics.helsinki.fi/courses/s/tl3/progs/slatecpackage.tgz

2) In files f90split.csh and slatec.csh change the appropriate string for the Fortran compiler: set FORT=gfortran (or f90, ...)

3) Split the one large file to smaller pieces and build the library: ./f90split.csh; ./slatec.csh

• Matlab has the function interp1 that does 1D interpolation using the method of choice (nearest, linear, spline, ...)

• Function spline computes the spline coefficients and they can be used to evaluate using function ppval: pp=spline(x,y) yi=ppval(pp,xi)

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Interpolation: practical tools

- A simple example using SLATEC:

program spline_slatec ixmax=(x2-x1)/dx implicit none ideriv=0 integer,parameter :: rk=selected_real_kind(15,100) inbv=1 real(rk),allocatable :: xa(:),ya(:),w(:),t(:),bcoef(:) do ix=1,ixmax real(rk) :: x,y,x1,x2,dx,fbcl,fbcr x=x1+dx*ix integer :: ibcl,ibcr,kntopt,ncoef,k,ideriv,inbv y=dbvalu(t,bcoef,n,k,ideriv,x,inbv,w) integer :: n,i,j,ix,ixmax write(6,*) x,y character(len=80) :: argu end do real(rk),external :: dbvalu stop if (iargc()/=3) then end program spline_slatec call getarg(0,argu) write(0,’(a,a,a)’) ’usage: ’,trim(argu),’ x1 x2 dx’ spline> ifort spline_slatec.f90 -lslatec -L../../slatec/ stop spline> a.out 1.0 8.0 0.01 < sin_per_1+x2.dat > end if sin_per_1+x2.interp call getarg(1,argu); read(argu,*) x1 call getarg(2,argu); read(argu,*) x2 call getarg(3,argu); read(argu,*) dx

read(5,*) n allocate(xa(1:n),ya(1:n),w(1:5*(n+2)),t(1:n+4),bcoef(1:n+2)) do i=1,n read(5,*) xa(i),ya(i) end do

ibcl=2 ibcr=2 fbcl=0.0 fbcr=0.0 kntopt=2 k=4 w=0.0

call dbint4(xa,ya,n,ibcl,ibcr,fbcl,fbcr,& & kntopt,t,bcoef,ncoef,k,w)

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