piecewise polynomial interpolationhermite interpolation 10 we can impose smoothness at the nodes to...

Piecewise Polynomial Interpolation

1

Piecewise linear interpolation

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Suppose we have data point (xk, yk), k = 0, 1, . . . N .

A piecewise linear polynomial that interpolates these

points is given by

p(x) = pk(x), x 2 [xk, xk+1]

where the polynomials pk(x) can be written as

pk(x) = a1(x� xk) + a0, k = 1, 2, . . . N � 1

At this point, we could easily compute the coe�cients

a = [a1, a0], but let’s go through a more general proce-

dure that we can use for higher order piecewise poly-

nomials.


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We can write this as a function of a local variable t which measures

the distance along the interval.

t =

x� xk

xk+1 � xkx = hkt+ xk

where hk = xk+1 � xk. Then, for 0 t 1, we have

pk(x) = epk(t) = ea1t+ ea0

(xk, yk)

(xk+1, yk+1)

t = 0

t = 1

0 < t < 1

epk(0) = yk

epk(1) = yk+1

Constraints to impose


4

This leads to a linear system

0 1

1 1

� ea1ea0

�=

ykyk+1

�

which we write as Aea = p. To solve for the coe�cients,

we use the matrix co-factor C to get A�1

A�1=

CT

det(A)

=

1

det(A)

1 �1

�1 0

�

We have det(A) = �1, and so

ea = �CTp


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To see what epk(t) looks like, we let q(t) = [t, 1] andwrite

epk(t) =

eaTq(t)= �pTCq(t)

= �⇥yk yk+1

⇤T

1 �1

�1 0

� t1

�

The advantage of this form is that we can clearly see

how the polynomial epk(t) depends on our data yk and

yk+1 :

epk(t) = (1� t)yk + tyk+1


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To build a spline in Matlab, we need to have the coef-

ficients a0 and a1 of our original polynomial.

pk(x) = a1(x� xk) + a0

We can recover these coe�cients by substituting t =

(x� xk)/hk into epk(t) = ea1t+ ea0 to get

pk(x) = ea1✓x� xk

hk

◆+ ea0 = a1(x� xk) + a0

where a1 = ea1/hk and ea0 = a0.


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These are exactly the coefficients that Matlab needs to build a spline using ‘mkpp’.

We can also write a = [a1, a0] using

a = �pTCS = �⇥yk yk+1

⇤ 1 �1

�1 0

� 1/hk 0

0 1

�

=

" yk+1 � ykhkyk

#

function pp_linear = build_linear_spline(xdata,ydata)N = length(xdata)-1;h = diff(xdata); %(x1-x0, x2-x1, x3-x2,...)C = [1 -1; -1 0];for k = 1:N, p = [ydata(k); ydata(k+1)]; S = diag([1./h(k) 1]); coeffs(k,:) = -p'*C*S;endpp_linear = mkpp(xdata,coeffs);end


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(x5, y5)

(x0, y0)

(x2, y2)

(x4, y4)


9

Can we come design a method that matches derivatives at the nodes as well?

Piecewise linear interpolation has these properties

• Continuous at nodes (xk, yk).

• Derivatives are not continuous (the function is

not smooth!)

Hermite interpolation

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We can impose smoothness at the nodes to get a nicer looking curve. Suppose we match the function values and derivatives.

p(xk) = yk

p

0(xk) = dk

p(xk+1) = yk+1

p

0(xk+1) = dk+1

Four conditions allow us to find the four coefficients of a cubic polynomial

This approach assumes that we have derivatives dk at each of the nodes xk.

pk(x) = a3(x� xk)3 + a2(x� xk)

2 + a1(x� xk) + a0


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As before, define

t =

x� xk

xk+1 � xk, ! x = hkt+ xk

where hk = xk+1 � xk. Then, for 0 t 1, we have

pk(x) ⌘ epk(t) = ea3t3 + ea2t2 + ea1t+ ea0

and our four conditions become

epk(0) = yk epk(1) = yk+1

ep0k(0) = dkhk ep0k(1) = dk+1hk


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This leads to the following linear system for the coe�cients ean :

We write the above system as Aea = p. To invert A, we use the

co-factor matrix C to get

A�1=

CT

det(A)

See Strang’s Introduction to Linear Algebra for more information on

the co-factor formula for the matrix inverse.


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We can then write epk(t) as

epk(t) = eaTq(t) = �pTCq(t)

where q(t) = [t3, t2, t, 1]T and we have used the fact that det(A) = �1.

Computing C, we get

epk(t) = �⇥yk yk+1 dkhk dk+1hk

⇤T

2

664

�2 3 0 �1

2 �3 0 0

�1 2 �1 0

�1 1 0 0

3

775

2

664

t3

t2

t1

3

775

which leads to a formula that clearly shows how the polynomial de-

pends on our data yk, yk+1, dk and dk+1:

epk(t) =

⇥1� t2(3� 2t)

⇤yk +

⇥t2(3� 2t)

⇤yk+1

+

⇥t(t� 1)

2⇤hkdk +

⇥t2(t� 1)

⇤hkdk+1

Piecewise Cubic Hermite Interpolating Polynomial (PCHIP)

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epk(t) =

⇥1� t

2(3� 2t)

⇤yk +

⇥t

2(3� 2t)

⇤yk+1

+

⇥t(t� 1)

2⇤hkdk +

⇥t

2(t� 1)

⇤hkdk+1

If we substitute t = (x � xk)/hk, we can recover our original poly-

nomial in x :

pk(x) = a3(x� xk)3+ a2(x� xk)

2+ a1(x� xk) + a0

where an = ean/hnk . We can write these coe�cients as

a = �pTCS

where S = diag(h

�3k , h

�2k , h

�1k , 1), p = [yk, yk+1, dkhk, dk+1hk], and

C is the co-factor matrix of our original coe�cient matrix A.

These are exactly the coefficients that Matlab needs to build a spline using ‘mkpp’.

The derivatives?

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Where do we get the derivatives dk from?

• If we know f(x), we can set dk = f

0(xk).

• We can approximate derivatives using data (xk, yk).

dk = G (�k�1, �k) , �k =yk+1 � yk

hk

where, for example, we could take G(a, b) = a+b2 , G(a, b) =

a, G(a, b) = b or some other reasonable function. Matlab’sPCHIP chooses derivatives to avoid introducing new extremain the spline.

• We can impose additional smoothness conditions and solveimplicitly for the derivatives.

Derivatives?

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Given data points (x0, y0), . . . , (xN , yN ), impose continuity in the

second derivative at the interior nodes,

p

00k(xk+1) = p

00k+1(xk+1), k = 1, 2, . . . N � 1.

This leads to a tridiagonal system for the derivatives at internal

nodes (”knots”) d1, d2, d3, . . . , dN�1.

To get equations for the end point derivatives, we will need to

to impose additional conditions at the endpoints.


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The entries in the tridiagonal system can be found by considering the polynomials

as functions of t :

p

00k�1(x) =

1

h

2k�1

p̃

00k�1(t), x = hk�1t+ xk�1

Transforming the equations equating second derivatives at knots, we obtain

1

h

2k�1

p̃

00k�1(1) =

1

h

2k

p̃

00k(0), k = 1, 2, 3, . . . N � 1

Using p̃

00k(t) = �pT

Cq00(t), we get

6

h

2k�1

yk�1 �6

h

2k�1

yk +2

hk�1dk�1 +

4

hk�1dk =

6

h

2k

yk �6

h

2k

yk+1 +2

hkdk +

4

hkdk+1

or

1

hk�1dk�1 +

✓1

hk�1+

1

hk

◆dk +

1

hkdk+1 =

3

h

2k�1

(yk � yk�1) +3

h

2k

(yk+1 � yk)

for k = 1, 2, 3, . . . , N � 1. This leads to a tridiagonal system for the unknown

derivatives dk.


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We can combine all equations into an (N �1)⇥ (N +1) ”augmented” tridiagonal

system A0d = b, or

2

666664

u0 u0 + u1 u1

u1 u1 + u2 u2

u2 u2 + u3 u3

.

.

.

.

.

.

.

.

.

uN�2 uN�2 + uN�1 uN�1

3

777775

2

666664

d0d1.

.

.

dN�1

dN

3

777775=

2

666664

b1.

.

.

.

.

.

bN�1

3

777775

where the matrix A0

A0=

⇥u0 A uN

⇤

and

ui =1

hi, k = 0, 1, 2, . . . , N � 1

bi = 3u2i�1(yi � yi�1) + 3u2

i (yi+1 � yi), k = 1, 2, 3, . . . , N � 1.

We can write this as a square tridiagonal system

Ad = b� d0u0 � dNuN

where u0 and uN are the first and last columns ofA0above, and d = (d1, d2, . . . , dN�1)


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We solve for the unknown derivatives d0, d1, d2, . . . , dN in three steps.

1. Solve dH = A�1b.

2. Solve

Ad0 = u0

AdN = uN

for vectors d0 and dN .

3. Solve

F(d0, dN ) =

a11 a12a21 a22

� d0dN

�+

F0

FN

�= 0

for scalars d0, dN so that so that ”endpoint” conditions (to be determined)

are satisfied. Coe�cients for the above system depend on choice of endpoint

conditions.

All unknown derivatives are then given by

d =

�d0, dH � d0d0 � dNdN , dN

End conditions?

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There are several additional conditions we can impose to solve for the endpoint

derivatives.

• Clamp derivative values d0 and dN to fixed values,

• Fit a cubic through first (last) four points and set d0 (dN ) to the derivative

of this polynomial at x0 (xN ).

• Impose continuity of the third derivative at x1 (xN�1). This is the ”not-a-

knot” condition, since we now have a single cubic polynomial over [x0, x2]

and [xN�2, xN ], e↵ectively eliminating the second (and second to last) node

(or ”knot”).

• For a periodic spline, match first and second derivatives between x0 and

xN .

• Impose the ’natural’ endpoint conditions, i.e. second derivative equal to 0.

Each of these can be represented as a 2⇥ 2 system of the form

F(d0, dN ) =

a11 a12

a21 a22

� d0

dN

�+

F0

FN

�= 0

After determining coe�cients, we can then solve for d0 and dN .

Cubic spline

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Periodic spline

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piecewise polynomial interpolationhermite interpolation 10 we can impose smoothness at the nodes to...

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