Bernstein polynomials

13 Pages·2000·0.082 MB·English

by  Kenneth I. Joy

On-LineGeometricModelingNotes
BERNSTEIN POLYNOMIALS
KennethI.Joy
VisualizationandGraphicsResearchGroup
DepartmentofComputerScience
UniversityofCalifornia,Davis
Overview
Polynomials are incredibly useful mathematical tools as they are simply defined, can be calculated
quickly on computer systems and represent a tremendous variety of functions. They can be differentiated
andintegratedeasily,andcanbepiecedtogethertoformsplinecurvesthatcanapproximateanyfunctionto
anyaccuracydesired. Moststudentsareintroductedtopolynomialsataveryearlystageintheirstudiesof
mathematics,andwouldprobablyrecallthemintheformbelow:
p(t) = a tn+a tn−1+···+a t+a
n n−1 1 0
whichrepresentsapolynomialasalinearcombinationofcertainelementarypolynomials(cid:8)(1,t,t2,...,tn(cid:9).
In general, any polynomial function that has degree less than or equal to n, can be written in this way,
andthereasonsaresimply
• The set of polynomials of degree less than or equal to n forms a vector space: polynomials can be
addedtogether,canbemultipliedbyascalar,andallthevectorspacepropertieshold.
• The set of functions (cid:8)1,t,t2,...,tn(cid:9) form a basis for this vector space – that is, any polynomial of
degreelessthanorequaltoncanbeuniquelywrittenasalinearcombinationsofthesefunctions.
This basis, commonly called the power basis, is only one of an infinite number of bases for the space of
polynomials.
Inthesenoteswediscussanotherofthecommonlyusedbasesforthespaceofpolynomials,theBernstein
basis,anddiscussitsmanyusefulproperties.
BernsteinPolynomials
TheBernsteinpolynomialsofdegreenaredefinedby
(cid:18) (cid:19)
n
B (t) = ti(1−t)n−i
i,n
i
fori = 0,1,...,n,where
(cid:18) (cid:19)
n n!
=
i i!(n−i)!
Therearen+1nth-degreeBernsteinpolynomials. Formathematicalconvenience,weusuallysetB = 0,
i,n
ifi < 0ori > n.
These polynomials are quite easy to write down: the coefficients (cid:0)n(cid:1) can be obtained from Pascal’s
i
triangle; the exponents on the t term increase by one as i increases; and the exponents on the (1−t) term
decreasebyoneasiincreases. Inthesimplecases,weobtain
• TheBernsteinpolynomialsofdegree1are
B (t) = 1−t
0,1
B (t) = t
1,1
andcanbeplottedfor0 ≤ t ≤ 1as
(cid:3)(cid:5)(cid:4)(cid:7)(cid:6)(cid:8)(cid:10)(cid:9)(cid:12)(cid:11)(cid:14)(cid:13) (cid:15)(cid:17)(cid:16)(cid:19)(cid:18)(cid:16)(cid:10)(cid:20)(cid:12)(cid:21)(cid:14)(cid:22)
(cid:2)
(cid:0) (cid:1)
2
• TheBernsteinpolynomialsofdegree2are
B (t) = (1−t)2
0,2
B (t) = 2t(1−t)
1,2
B (t) = t2
2,2
andcanbeplottedfor0 ≤ t ≤ 1as
(cid:3)(cid:5)(cid:4)(cid:7)(cid:6)(cid:8)(cid:10)(cid:9)(cid:12)(cid:11)(cid:14)(cid:13) (cid:24)(cid:26)(cid:25)(cid:28)(cid:27)(cid:25)(cid:10)(cid:29)(cid:12)(cid:30)(cid:14)(cid:31)
(cid:15)(cid:17)(cid:16)(cid:19)(cid:18)(cid:20)(cid:10)(cid:21)(cid:12)(cid:22)(cid:14)(cid:23)
(cid:2)
(cid:0) (cid:1)
• TheBernsteinpolynomialsofdegree3are
B (t) = (1−t)3
0,3
B (t) = 3t(1−t)2
1,3
B (t) = 3t2(1−t)
2,3
B (t) = t3
3,3
andcanbeplottedfor0 ≤ t ≤ 1as
3
(cid:3)(cid:5)(cid:4)(cid:7)(cid:6)(cid:8)(cid:10)(cid:9)(cid:12)(cid:11)(cid:14)(cid:13) !(cid:26)"(cid:28)#"(cid:10)$(cid:12)%(cid:14)&
(cid:15)(cid:17)(cid:16)(cid:19)(cid:18)(cid:20)(cid:10)(cid:21)(cid:12)(cid:22)(cid:14)(cid:23) (cid:24)(cid:26)(cid:25)(cid:28)(cid:27)(cid:29)(cid:10)(cid:30)(cid:12)(cid:31)(cid:14)
(cid:2)
(cid:0) (cid:1)
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ARecursiveDefinitionoftheBernsteinPolynomials
TheBernsteinpolynomialsofdegreencanbedefinedbyblendingtogethertwoBernsteinpolynomials
ofdegreen−1. Thatis,thekthnth-degreeBernsteinpolynomialcanbewrittenas
B (t) = (1−t)B (t)+tB (t)
k,n k,n−1 k−1,n−1
Toshowthis,weneedonlyusethedefinitionoftheBernsteinpolynomialsandsomesimplealgebra:
(cid:18) (cid:19) (cid:18) (cid:19)
n−1 n−1
(1−t)B (t)+tB (t) = (1−t) tk(1−t)n−1−k +t tk−1(1−t)n−1−(k−1)
k,n−1 k−1,n−1
k k−1
(cid:18) (cid:19) (cid:18) (cid:19)
n−1 n−1
= tk(1−t)n−k + tk(1−t)n−k
k k−1
(cid:20)(cid:18) (cid:19) (cid:18) (cid:19)(cid:21)
n−1 n−1
= + tk(1−t)n−k
k k−1
(cid:18) (cid:19)
n
= tk(1−t)n−k
k
= B (t)
k,n
TheBernsteinPolynomialsareAllNon-Negative
A function f(t) is non-negative over an interval [a,b] if f(t) ≥ 0 for t ∈ [a,b]. In the case of the
Bernstein polynomials of degree n, each is non-negative over the interval [0,1]. To show this we use the
recursivedefinitionpropertyaboveandmathematicalinduction.
ItiseasilyseenthatthefunctionsB (t) = 1−tandB (t) = tarebothnon-negativefor0 ≤ t ≤ 1. If
0,1 1,1
weassumethatallBernsteinpolynomialsofdegreelessthankarenon-negative,thenbyusingtherecursive
definitionoftheBernsteinpolynomial,wecanwrite
B (t) = (1−t)B (t)+tB (t)
i,k i,k−1 i−1,k−1
and argue that B (t) is also non-negative for 0 ≤ t ≤ 1, since all components on the right-hand side
i,k
of the equation are non-negative components for 0 ≤ t ≤ 1. By induction, all Bernstein polynomials are
non-negativefor0 ≤ t ≤ 1.
Inthisprocess,wehavealsoshownthateachoftheBernsteinpolynomialsispositivewhen0 < t < 1.
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TheBernsteinPolynomialsformaPartitionofUnity
Asetoffunctionsf (t)issaidtopartitionunityiftheysumtooneforallvaluesoft. Thek+1Bernstein
i
polynomialsofdegreek formapartitionofunityinthattheyallsumtoone.
To show that this is true, it is easiest to first show a slightly different fact: for each k, the sum of the
k+1BernsteinpolynomialsofdegreekisequaltothesumofthekBernsteinpolynomialsofdegreek−1.
Thatis,
k k−1
X X
B (t) = B (t)
i,k i,k−1
i=0 i=0
Thiscalculationisstraightforward,usingtherecursivedefinitionandcleverlyrearrangingthesums:
k k
X X
B (t) = [(1−t)B (t)+tB (t)]
i,k i,k−1 i−1,k−1
i=0 i=0
"k−1 # " k #
X X
= (1−t) B (t)+B (t) +t B (t)+B (t)
i,k−1 k,k−1 i−1,k−1 −1,k−1
i=0 i=1
k−1 k
X X
= (1−t) B (t)+t B (t)
i,k−1 i−1,k−1
i=0 i=1
k−1 k−1
X X
= (1−t) B (t)+t B (t)
i,k−1 i,k−1
i=0 i=0
k−1
X
= B (t)
i,k−1
i=0
(wherewehaveutilizedB (t) = B (t) = 0).
k,k−1 −1,k−1
Oncewehaveestablishedthisequality,itissimpletowrite
n n−1 n−2 1
X X X X
B (t) = B (t) = B (t) = ··· = B (t) = (1−t)+t = 1
i,n i,n−1 i,n−2 i,1
i=0 i=0 i=0 i=0
The partition of unity is a very important property when utilizing Bernstein polynomials in geometric
modeling and computer graphics. In particular, for any set of points P , P , ..., P , in three-dimensional
0 1 n
space,andforanyt,theexpression
P(t) = P B (t)+P B (t)+···+P B (t)
0 0,n 1 1,n n n,n
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isanaffinecombinationofthesetofpointsP ,P ,...,P andif0 ≤ t ≤ 1, itisaconvexcombinationof
0 1 n
thepoints.
DegreeRaising
Anyofthelower-degreeBernsteinpolynomials(degree<n)canbeexpressedasalinearcombination
ofBernsteinpolynomialsofdegreen. Inparticular,anyBernsteinpolynomialofdegreen−1canbewritten
asalinearcombinationofBernsteinpolynomialsofdegreen. Wefirstnotethat
(cid:18) (cid:19)
n
tB (t) = ti+1(1−t)n−i
i,n
i
(cid:18) (cid:19)
n
= ti+1(1−t)(n+1)−(i+1)
i
(cid:0)n(cid:1)
= i B (t)
(cid:0)n+1(cid:1) i+1,n+1
i+i
i+1
= B (t)
i+1,n+1
n+1
and
(cid:18) (cid:19)
n
(1−t)B (t) = ti(1−t)n+1−i
i,n
i
(cid:0)n(cid:1)
= i B (t)
(cid:0)n+1(cid:1) i,n+1
i
n−i+1
= B (t)
i,n+1
n+1
andfinally
1 1
B (t)+ B (t) = ti(1−t)n−i+ti+1(1−t)n−(i+1)
(cid:0)n(cid:1) i,n (cid:0) n (cid:1) i+1,n
i i+1
= ti(1−t)n−i−1((1−t)+t)
= ti(1−t)n−i−1
1
= B (t)
(cid:0)n−1(cid:1) i,n−1
i
Usingthisfinalequation,wecanwriteanarbitraryBernsteinpolynomialintermsofBernsteinpolynomials
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ofhigherdegree. Thatis,
" #
(cid:18) (cid:19)
n−1 1 1
B (t) = B (t)+ B (t)
i,n−1 i (cid:0)n(cid:1) i,n (cid:0) n (cid:1) i+1,n
i i+1
(cid:18) (cid:19) (cid:18) (cid:19)
n−i i+1
= B (t)+ B (t)
i,n i+1,n
n n
which expresses a Bernstein polynomial of degree n − 1 in terms of a linear combination of Bernstein
polynomials of degree n. We can easily extend this to show that any Bernstein polynomial of degree k
(lessthann)canbewrittenasalinearcombinationofBernsteinpolynomialsofdegreen–e.g.,aBernstein
polynomialofdegreen−2canbeexpressedasalinearcombinationoftwoBernsteinpolynomialsofdegree
n−1, each of which can be expressed as a linear combination of two Bernstein polynomials of degree n,
etc.
ConvertingfromtheBernsteinBasistothePowerBasis
Sincethepowerbasis{1,t,t2,...,tn}formsabasisforthespaceofpolynomialsofdegreelessthanor
equal to n, any Bernstein polynomial of degree n can be written in terms of the power basis. This can be
directlycalculatedusingthedefinitionoftheBernsteinpolynomialsandthebinomialtheorem,asfollows:
(cid:18) (cid:19)
n
B (t) = tk(1−t)n−k
k,n
k
(cid:18) (cid:19) n−k (cid:18) (cid:19)
n X n−k
= tk (−1)i ti
k i
i=0
n−k (cid:18) (cid:19)(cid:18) (cid:19)
X n n−k
= (−1)i ti+k
k i
i=0
n (cid:18) (cid:19)(cid:18) (cid:19)
X n n−k
= (−1)i−k ti
k i−k
i=k
n (cid:18) (cid:19)(cid:18) (cid:19)
X n i
= (−1)i−k ti
i k
i=k
wherewehaveusedthebinomialtheoremtoexpand(1−t)n−k.
ToshowthateachpowerbasiselementcanbewrittenasalinearcombinationofBernsteinPolynomials,
8
weusethedegreeelevationformulasandinductiontocalculate:
tk = t(tk−1)
n (cid:0) i (cid:1)
X
= t k−1 B (t)
(cid:0) n (cid:1) i,n−1
i=k−1 k−1
n (cid:0)i−1(cid:1)
X
= k−1 tB (t)
(cid:0)n−1(cid:1) i−1,n−1
i=k k−1
n−1 (cid:0) i (cid:1)
X i
= k−1 B (t)
(cid:0) n (cid:1)n i,n
i=k−1 k−1
n−1 (cid:0)i(cid:1)
X
= k B (t),
(cid:0)n(cid:1) i,n
i=k−1 k
wheretheinductionhypothesiswasusedinthesecondstep.
Derivatives
DerivativesofthenthdegreeBernsteinpolynomialsarepolynomialsofdegreen−1. Usingthedefini-
tionoftheBernsteinpolynomialwecanshowthatthisderivativecanbewrittenasalinearcombinationof
Bernsteinpolynomials. Inparticular
d
B (t) = n(B (t)−B (t))
k,n k−1,n−1 k,n−1
dt
for0 ≤ k ≤ n. Thiscanbeshownbydirectdifferentiation
(cid:18) (cid:19)
d d n
B (t) = tk(1−t)n−k
k,n
dt dt k
kn! (n−k)n!
= tk−1(1−t)n−k + tk(1−t)n−k−1
k!(n−k)! k!(n−k)!
n(n−1)! n(n−1)!
= tk−1(1−t)n−k + tk(1−t)n−k−1
(k−1)!(n−k)! k!(n−k−1)!
(cid:18) (cid:19)
(n−1)! (n−1)!
= n tk−1(1−t)n−k + tk(1−t)n−k−1
(k−1)!(n−k)! k!(n−k−1)!
= n(B (t)−B (t))
k−1,n−1 k,n−1
9
Thatis,thederivativeofaBernsteinpolynomialcanbeexpressedasthedegreeofthepolynomial,multiplied
bythedifferenceoftwoBernsteinpolynomialsofdegreen−1.
TheBernsteinPolynomialsasaBasis
Why do the Bernstein polynomials of order n form a basis for the space of polynomials of degree less
thanorequalton?
1. Theyspanthespaceofpolynomials–anypolynomialofdegreelessthanorequaltoncanbewritten
asalinearcombinationoftheBernsteinpolynomials.
ThisiseasilyseenifonerealizesthatThepowerbasisspansthespaceofpolynomialsandanymember
ofthepowerbasiscanbewrittenasalinearcombinationofBernsteinpolynomials.
2. Theyarelinearlyindependent–thatis,ifthereexistconstantsc ,c ,...,c sothattheidentity
0 1 n
0 = c B (t)+c B (t)+···+c B (t)
0 0,n 1 1,n n n,n
holdsforallt,thenallthec ’smustbezero.
i
Ifthisweretrue,thenwecouldwrite
0 = c B (t)+c B (t)+···+c B (t)
0 0,n 1 1,n n n,n
n (cid:18) (cid:19)(cid:18) (cid:19) n (cid:18) (cid:19)(cid:18) (cid:19) n (cid:18) (cid:19)(cid:18) (cid:19)
X n i X n i X n i
= c (−1)i ti+c (−1)i−1 ti+···+c (−1)i−n ti
0 1 n
i 0 i 1 i n
i=0 i=1 i=n
" 1 (cid:18) (cid:19)(cid:18) (cid:19)# " n (cid:18) (cid:19)(cid:18) (cid:19)#
X n 1 X n n
= c + c t1+···+ c tn
0 i i
1 1 n n
i=0 i=0
Sincethepowerbasisisalinearlyindependentset,wemusthavethat
c = 0
0
1 (cid:18) (cid:19)(cid:18) (cid:19)
X n 1
c = 0
i
1 1
i=0
.
.
.
n (cid:18) (cid:19)(cid:18) (cid:19)
X n n
c = 0
i
n n
i=0
whichimpliesthatc = c = ··· = c = 0(c isclearlyzero,substitutingthisinthesecondequation
0 1 n 0
10