ch5_4.doc'kfxzghj,

7/27/2019 ch5_4.doc'kfxzghj,

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Chapter 5. B-Splines

5.1. First Order B-Spline

B-splines (think Basis splines) form a truly piecewise basis for the spline family

as can be learned later. For a set of infinite data points ),,( =ixi where 1+< ii xx

for any i , the B-splines of degree k are defined as

1),()()( 1111

11

+

= ++++

++

+

kxBxx

xxxB

xx

xxxB ki

iki

kik

i

iki

ik

i (5.1.1)

and

=

== +

otherwise

xxxxB

ii

i,0

),[,1)(

10(5.1.2)

where i and kare integers as usual. Note: 0)(1 xBk

i always.Each )(

0 xBi is nonzero in exactly one interval (Figure 5.1.1) and has a

discontinuity at 1+ix . For any dataset ),,2,1,,( niyx ii = , the constant B-spline is

=

=n

i

ii xByxs1

0 )()( (5.1.3)

which simply extends each iy across eachthi interval.

x(i-1) x(i) x(i+1) x(i+2) x(i+3) x(i+4)0

0.5

1

1.5

Bi0 B

i+20

Figure 5.1.1 Non-zero parts of B-splines of degree 0.

1


2/18

For 1=k , )(1xBi can be derived from Equations 5.1.1.and 5.1.2 as

=

+

=

++++

+

+

+

+++

+

+

elsewhere

xxxxx

xx

xxx

xx

xx

xBxx

xxxB

xx

xxxB

ii

ii

i

ii

ii

i

i

ii

ii

i

ii

i

i

,0

),[,

),[,

)()()(

21

12

2

1

1

0

1

12

20

1

1

(5.1.4)

which has two non-zero parts, in intervals [ )1, + ii xxx and [ )21 , ++ ii xxx as shown inFigure 5.1.2 (Dashed lines are used in order to make the splines more distinguishable).

These intervals are called the support of )(1 xBi . On each of its support intervals,

)(1 xBi is a linear function and its location and slope are solely determined by the

distribution of the ix s. )(1 xBi has a peak value of 1 at 1+ix and is continuous there. It

is quite clear that )(1xBi has

0C but not 1C continuity.

2


3/18

x(i-1) x(i) x(i+1) x(i+2) x(i+3) x(i+4)0

0.5

1

1.5

Bi-1

1 Bi

1 Bi+1

1 Bi+2

1

Figure 5.1.2 Non-zero parts of B-splines of degree 1.

It is interesting to note that )(1 xBi has some connection to the elementary

Lagrange interpolating polynomial. If we write the linear Lagrange interpolating

polynomial for the two-point data set of )1and,( += iijxj , the two elements would be:

ii

i

ii

i

xx

xxxl

xx

xxxl

=

=

+++

1

22

1

121 )(,)(

which are the second part of )(1

1 xBi and the first part of )(1xBi , respectively. As we

shall learn, this agreement does not extend to higher order B-splines.

In general, the 1st order B-spline is constructed from

)()( 1 1 xBDxs ii

i

=

= (5.1.5)

where iD are constants determined from the data. However, we know from Section 4.1.1

that the linear spline on a given data interval is also the Lagrange interpolation formula

for the data set consisting of the two end points of the same data interval. So lets discuss

the relationship of )(1 xBi with the linear splines. It turns out that for any data set

)infinitybecanwhere,,,2,1,,( nniyx ii = , the following function is a linear spline:

3

Support intervals for


4/18

)()(1

1

1

xByxs i

n

i

i

=

= (5.1.6)

To prove it, lets look at the formula of )(xs on interval [ ]1, +ii xx . Of all the

terms in the above equation, only)(

1

1

xBi and

)(1xB

i have non-zero contributions onthe interval. Thus

],[,)()()( 11

1

1

1 ++ += iiiiii xxxxByxByxs (5.1.7)

On the interval [ ]1, +ii xx , )(1 1 xBi will take the second leg of its formula and)(1 xBi will take the first leg of its formula (see Equation 5.1.4). Thus

],[,)( 11

1

1

1

++

++

+

+

= iiii

i

i

ii

i

i xxxxx

xxy

xx

xxyxs (5.1.8)

which is exactly the linear spline (see Equation 4.1.1) as well as the linear Lagrange

interpolating polynomial.

Say 2211 ,;, yxyx are the only nonzero data, then

0)()(0)(1

12

1

01+++++= xByxByxs

where

[ )

[ )

=

21

12

2

10

01

0

10

,

,

)(

xxxxx

xx

xxx

xx

xx

xB

and

[ )

[ )

=

32

23

3

21

12

1

1

1

,

,

)(

xxxxxxx

xxxxx

xx

xB

4


5/18

[ )

[ )

[ )

+

=

extensionRight,x

LagrangeisrowThis,x

extensionLeft,x

)(

32

32

32

21

12

12

21

21

10

01

01

xxxx

xxy

xxxx

xx

yxx

xx

y

xxxx

xxy

xs

The central row is the same as a Lagrange interpolating polynomial, but )(xs in

terms of B-splines is more general: it contains Lagrange plus two arbitrary end extensions

(points 30 ,xx are not part of the data set) as seen in Fig 5.1.3.

Now we see why we would use )(1 1 xBi instead of )(1 xBi in Equation 5.1.5

and 5.1.6. The reason is that the hat shape of )(1 xBi (Figure 5.1.2) is centered at 1+ix ,

not ix . For iy s contribution to take place, it should be teamed with a component of B-

splines

Figure 5.1.3 First degree B-spline for two data points.

that lags by 1. So yi is teamed with)(1

1

xBi . Similar notation is used for all higher order

B-splines. For example, in the case of 2=k which will be discussed later, the sum overi of terms iy )(

2

2 xBi are used together. Furthermore, since )(1

1 xBi is centered on

ix , ii yD = . Further insight can be gleamed from Figure 5.1.2.

It is worth noting that on [ )1, + ii xxx , )(1 1 xBi + )(1 xBi = 1 exactly. As wewill prove later, this is true for all orders of B-splines on all intervals.

5

0

0B0

1B0

2B

1

y1

y2

xx

0x

1 x2 x3


6/18

Generally for linear splines, two points are used to draw a straight line for each

interval. Then for each data point, its x and y information will contribute to two splines.

In the case of )(1xBi , this contribution is built into its expression through its two legs.

This is the essence of all B-splines and one reason why they are called basis splines.

Polynomial splines and Lagrange interpolating polynomials discussed before can

all be viewed as functions over interval(s), as functions should be. The emphasis here ison the intervals. A polynomial is set up over the interval(s), then, we go to its end points

for help on determining the coefficients of the polynomials.

B-splines, however, should be viewed more as belonging to the points. Each

point function covers certain interval(s). To build splines through B-splines, we will

always use the form similar to that in Equation 5.1.5. The y information is applied to thepoint function and we are done (almost, anyway) with the building of the splines. This

feature will be more evident in the case for )(2 xBi .

6


7/18

5.2 Second Order B-Splines

Now for 2=k , from Equation (5.1.1) and (5.1.2), we have

)()()( 1 113

31

2

2 xBxx

xxxB

xx

xxxB i

ii

ii

ii

ii

+++

+

+

+

=

+

= +++

+

++

)()(0

1

12

20

12

xBxx

xxxB

xx

xx

xx

xxi

ii

i

i

ii

i

ii

i

+

+ +++

++

++

+

++

+ )()(0

2

23

30

1

12

1

13

3 xBxx

xxxB

xx

xx

xx

xxi

ii

i

i

ii

i

ii

i

+

=

++

++++

+

++

++

++

+

+

++

+

++

elsewhere

xxx

xxxx

xx

xxx

xx

xxxx

xx

xxxx

xx

xxxxxxx

xx

xB

ii

iiii

i

ii

ii

ii

ii

ii

ii

ii

iiii

i

i

,0

),[,))((

)(

),[,))(())((1

),[,))((

)(

)(

32

2313

2

3

21

13

31

2

2

12

1

21

2

2(5.2.1)

Non-zero parts of three2

B splines are plotted in Figures 5.2.1. The support ofeach spline covers three intervals. It is a quadratic polynomial on each support interval.

Note that the peak value of )(2 xBi is less than 1. This is different from )(

0 xBi , )(1 xBi

and other interpolation functions that do have peak values of 1. If all the 2B splines

could be plotted, in any interval, there would be contributions from three splines. But we

only see this in ],[ 21 ++ ii xx in Figure 5.2.1. Interestingly, these three contributions sum

to unity.

7


8/18

Also, the components of )(2 xBi differ from the quadratic Lagrange polynomials.

For comparison, Figure 5.2.2 shows three quadratic elementary Lagrange polynomials for

the data set of )2and1,,( ++= iiijxj . The3

1l term goes negative. Different indeed!

x(i-1) x(i) x(i+1) x(i+2) x(i+3) x(i+4)0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Bi-12

Bi2

Bi+12

Figure 5.2.1 Non-zero parts of B-splines of degree 2. Note that the peak values are less than 1, but the

splines in ],[ 21 ++ ii xx sum to 1.

x(i-1) x(i) x(i+1) x(i+2) x(i+3) x(i+4)-

- .

0.5

1

1.5

3(x)

3(x)

l3(x)

8


9/18

Figure 5.2.2 Comparison between B-spline of degree 2 and the elementary quadratic Lagrange

polynomials. Note: one component goes negative, however =

=3

1

3 1i

il .

By taking derivatives of all the sections in Equation 5.2.1, we get

++

+

=

++

++++

+

++

++

++

+

+

++

+

++

elsewhere

xxxxxxx

xx

xxx

xx

xxxx

xx

xxxx

xx

xxxxxxx

xx

xB

ii

iiii

i

ii

ii

ii

ii

ii

ii

ii

iiii

i

i

,0

),[,))((

)(2

),[,1

),[,))((

)(2

)(

32

2313

3

21

13

13

2

2

12

1

21

2

9


10/18

=

=

=

=

=

=

=

=

=

+

++++++

++++

++

+

+

++

+++

++

pointsdataotherallat0

0)(,2

))((

)(2)(

2)(,

2)(

2

))((

)(2)(,0)(

2

3

2

132313

232

2

13

2

2

2

1

2

221

11

22

i

ii

iiiiii

ii

ii

ii

ii

ii

ii

iiiiii

ii

iiii

B

xBxxxxxx

xxxB

xxxB

xxxB

xxxxxx

xxxBxB

(5.2.2)

The second part of line one in Equation 5.2.2 is exactly the first part of line two,

so2

iB is continuous at 1+ix . The second part of line two is also exactly the first part of

line three, thus2

iB is continuous at 2+ix . The first part of line one equals line four and

2iB is continuous at ix . The second part of line three equals line four and

2iB is

continuous at 3+ix . Line 4 also means2

iB is continuous at all the other data points.

So,2

iB is continuous at all the data points. From Equation 5.2.1, we know2

iB

is continuous at all the locations between the data points. Thus2

iB is continuous at any

x and )(2 xBi has1C continuity. Any linear combination of )(

2 xBi s also has1C

continuity.

It can be shown that for any k

=

=i

k

i xB 1)( (5.2.3)

10


11/18

For 0=k and for any x , some m can be found such that [ )1, + mm xxx . Of allthe )(

0xBi with = ,i , only )(

0 xBm will be nonzero, thus

=

==i

mi xBxB 1)()(00

For 0>k , consider a function f which is a linear combination of )(xB ki

=

=i

k

i

k

i xBDxf )()(

Note here that the proof of Equation 5.2.3 does not requirek

iD to be arbitrary, but rather

1=kiD is enough.k

iD is used to illustrate its relationship with lower order terms.

=

+

+++

++

+

+

= ik

i

iki

kik

i

iki

ik

i xBxx

xxxBxx

xxDxf )()()(

1

1

11

11

=

+

+++

++

=

+

+

=

i

k

i

iki

kik

i

i

k

i

iki

ik

ixB

xx

xxDxB

xx

xxD )()(

1

1

11

11

Denoting 1+= ij , the second term above can be rewritten as

=

+

+

=

+

+

+

+=

+

+

=

=

i

k

i

iki

kik

i

j

k

j

jkj

kjk

j

j

k

j

jkj

kjk

j xBxx

xxDxB

xx

xxDxB

xx

xxD )()()( 11

1

1

1

1

1

1

So, f can be rewritten as

=

+

+

=

+

+

=

i

k

i

iki

kik

i

i

k

i

iki

ik

ixB

xx

xxDxB

xx

xxDxf )()()(

1

1

1

=

+

+

+

+

=

i

k

i

iki

kik

i

iki

ik

i xBxx

xxD

xx

xxD )(11

Introducing

iki

kik

iiki

ik

i

k

i xx

xxD

xx

xxDD

+

=

+

+

+

1

1

(5.2.4)and we have

=

=

i

k

i

k

i xBDxf )()(11

11


12/18

which means any function f that can be written as a linear combination of )(xBk

i can

also be written as a linear combination of )(1xB

k

i

. Equation 5.2.4 is the relationship

between their respective coefficients.

The above equations are recursive, )(xf can be written as combinations of

)(,),(),(021xBxBxB i

k

i

k

i

while k is decreased to 0,,2,1 kk . In the case k

is eventually reduced to 0, we get

=

=i

iixBDxf )()( 00

and0

iD can be calculated using the recursive scheme starting fromk

iD . So, ifk

iD are

known for all = ,i ,0

iD would be known for all = ,i .

Now, considering a specific function )(xf , where itsk

iD = 1 for all = ,i ,

the goal here is to find out what is )(xf exactly. In Equation 5.2.3, it is stated that)(xf is exactly 1. Now we can prove that.

From Equation 5.2.4, we get for any i

11

1 =

+

=

+

=

+

+

++

+

+

iki

ki

iki

i

iki

kik

i

iki

ik

i

k

ixx

xx

xx

xx

xx

xxD

xx

xxDD

11

1

11

11

1

1

12 =

+

=

+

=

+

+

++

+

+

iki

ki

iki

i

iki

kik

i

iki

ik

i

k

ixx

xx

xx

xx

xx

xxD

xx

xxDD

Repeating this process, we eventually get

11

1

11

11

1

1

10 =

+

=

+

=

+

+

+++

++

++ ii

i

ii

i

ikki

kki

i

ikki

i

iixx

xx

xx

xx

xx

xxD

xx

xxDD

which means

1)()()( 000

=

=

===i

i

i

iixBxBDxf

where at the second step in the above equation, the result for 0=k is used. Thus,Equation 5.2.3 has been proven.

Example 5.2.1 Illustrative example to demonstrate Equation 5.2.3

Say [ ]10 ,xxx , then

12


13/18

[ ]

[ ]

1

)(

1

)(

)(

)(

)(

))((

)(

))((

)(

))((

)()(

))((

)(

))((

))((

))((

)(

)(

))((

)(

)(

))((

)()(

))((

))((

)(

))((

)(

0

)()()()(

0

)()(

01

01

01

0

01

1

02

0102

0

0102

2

0

0102

02

11

0111

1

0111

11

0111

2

1

2

0

0102

2

0

2

1

0102

02

0111

11

2

2

0111

2

1

2

1

2

0

2

1

2

2

2

3

2

=

+

=

+

=

+

+

+

+

+

=

+

=

+

+

+

+

+

=

++++++=

=

xxxxxx

xx

xx

xx

xx

xxxxxxxx

xx

xxxx

xx

xxxx

xxxx

xxxxxxxx

xx

xxxx

xxxx

xxxx

xx

xB

xxxx

xx

xB

xxxx

xxxx

xxxx

xxxx

xB

xxxx

xx

xBxBxBxBxBxBi

i

5.3. Building Second Order B-Splines

Considering a data set: ( )niyx ii ,,2,1,, = , we wish to find the B-spline of

the 2nd

order. First, write the spline function as a linear combination of )(2

xBi

=

=i

iixBDxs )()(

2

2 (5.3.1)

where the index i goes from to in general and is consistent with what we usein Section 5.1. We should view the data set ( )niyx ii ,,2,1,, = as part of a data setthat has an infinite number of knots. As will be seen later, the rest of the infinite data set

is of no use to us for the current task. Also note, the subscripts on 2B are shifted, e.g.,

)(2

2xBD ii is used here for the same reason mentioned for the )(

1 xBi case.

As stated before, the support of this spline covers only three intervals. Soto find the splines that have contributions in the data set range, we need to find the left

most spline that has contributions to the 1st data interval and the right most spline that

contributes to the last data interval. As seen from the Figure 5.3.1, they should be

)(21 xB and )(2

1 xBn , respectively. So the contributing B2s are

2

1

2

0

2

1 ,,, nBBB .

Hence the required range for 2iiBD is 11 + ni . Equation 5.3.1 can then berewritten as

13


14/18

+

==

1

1

2

2 )()(n

i

ii xBDxs(5.3.2)

Figure 5.3.1.

The number of iD to be solved is 1+n , which is one more than the number of

knots (or data points). Right away, we know the 0C continuity can give us n conditionsand one more condition is needed.

To apply the 0C continuity conditions, we need to calculate the value of )(xs at

the knots. Looking at point ix in Figure 5.3.1, it is clear that only )(2

2 xBi and

)(2

1 xBi are nonzero. Thus

iiiiiiii yxBDxBDxs += + )()()(2

11

2

2 (5.3.3)

14

x1

x0

x2

x-1

xn-2

xn-1

xn

xn+1

Out of rangeOut of range

Data set

2

1B2

1nB


15/18

From Equation 5.2.1, setting 2 ii , and letting iii xxh = +1 , we get

)(

)(

))((

)()(

1

11

1

111

2

12

2

++

++

+

+

=

=

i

ii

i

ii

ii

iiii

iiii

h

xx

h

xx

xx

xxxx

xxxB

1

2

2 )(

+=

ii

i

iihh

hxB (5.3.4)

Similarly setting 1 ii in Equation 5.2.1 gives

)(

)(

)(

))((

)(

))((

)(

1)(

11

1

2

2

11

11

1

2

2

+

+

+

+

+

+

=

+

=

ii

ii

ii

iiii

ii

iiii

ii

ii

xx

xx

xx

xxxx

xx

xxxx

xxxB

1

12

2 )(

+

=ii

iii

hh

hxB (5.3.5)

Hence from Equation 5.3.3 - 5, we get

i

ii

i

i

ii

i

i yhh

hD

hh

hD =

++

+

+ 1

1

1

1

iiiiiii yhhDhDh )( 111 + +=+ (5.3.6)

where iii xxh = +1 . Equation 5.3.6 contains n equations and there are 1+n unknowns.Certainly, we can assign a value to 1D and use the above equation as a recursive scheme

to calculate all the rest of the coefficients. The value of 1D can be zero, or can be

calculated by some other methods (e.g., Cheney and Kincaid, 1994).

The values of 0x and 1+nx need to be assigned. If we define h-ratio11

0 :

=n

n

h

h

h

h

, then these can be set by the user. Usually they are assigned such that h-ratio 1:1= .

Example 5.3.1. Given 3 data points ( )3,2,1,, =iyx ii , solve for constants D.

15


16/18

( )

( )

( )301

3234233

2123122

1012011

,,givenunknowns3inequations3

:3

:2

:1

hhD

yhhDhDhi

yhhDhDhi

yhhDhDhi

+=+=

+=+=

+=+=

To solve, we must choose

Constraint like 01 =D or other.

To find 30 ,hh , vary the h-ratios2

3

1

0 ,h

h

h

hthat, in concert with 4,,1, =iDi ,

yield tautness to the spline.

The algorithm of the MATLAB code bspline2.m follows

(1) Input x where the interpolation value )(xsy = is needed;(2) Check to see ifx is out of range; if yes, abort the program; if no, proceed;(3) Calculate or assign 1D (See Cheney and Kincaid, 1994);

(4) Calculate iD (Equation 5.3.3);

(5) Find the data interval where x belongs;(6) Calculate y according to Equation 5.3.2.

Example 5.3.2. Show 2nd order B-splines for data sets in Examples 4.1.1 and 4.1.2

Figures 5.3.1 and 5.3.2 show the 2nd order B-splines for the data sets in Examples

4.1.1 and 4.1.2, respectively. Three sets of 0h and nh are used.

16
http://www.esm.psu.edu/courses/emch407/njs/m-files02/bspline2.mhttp://www.esm.psu.edu/courses/emch407/njs/m-files02/bspline2.m


17/18

0 1 2 3 4 5 6 7 8 9 10-5

0

5

10

15

20

x

y

Quadratic B-Spline

d* => d calculated perCheney and Kincaid (1994)

h-ratio = 100:1 and d*

h-ratio = 1:1 and d = 0



Figure 5.3.1 Second Order B-spline for dataset from Example 4.1.1.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

x

y

Quadratic B-Spline

h-ratio = 1:0.1 and d*



h-ratio = 1:1 and d = 0

d* => d calculated per

Cheney and Kincaid (1994)

Figure 5.3.2 Second Order B-spline for data set from Example 4.1.2.

17


18/18

We know )(xs in Equation 5.3.2 has 0C continuity, but what about 1C

continuity? Recalling from Section 4.2, the quadratic spline has 1C continuity. What

about )(xs in Equation 5.3.2? The answer is yes. Since )(xs is a linear combination of

the )(2xBi s which have

1C continuity, )(xs has 1C continuity too.

In general, )(xBk

i has continuous derivatives up to order 1k (see Cheney and

Kincaid, 1994).

Conclusion

In conclusion, we discovered that the B-splines:

(1) Have 0C continuity for any order.

(2) The first order spline requires no information in addition to the data points.

(3) The second order spline requires three additional pieces of information (one more

interval on each end of the data set and, say, 1d ) regardless of the number of data

points.

(4) The second order spline has1

C continuity.

Reference

Cheney, W. and Kincaid, D., Numerical Mathematics and Computing, Third Edition,Brooks/Cole Publishing Company, 1994.

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