Approximation and Bernstein Polynomials

8/13/2019 Approximation and Bernstein Polynomials

1/16


2/16

W. Qian et al. / European Journal of Combinatorics 32 (2011) 448463 449

The Weierstrass Approximation Theorem. Let f be a continuous function defined on the closed interval

[a, b]. For any >0, there exists a polynomial function p such that for all x in [a, b], we have

|f(x)p(x)|< .

The theorem can be proved by a transformation with Bernstein polynomials [8]. By a linearsubstitution, the interval [a, b] can be transformed into the unit interval [0, 1]. Thus, the originalstatement of the theorem holds if and only if the theorem holds for every continuous function fdefined on the interval[0, 1].

ABernstein polynomialof degreenis a polynomial expressed in the following form [1]:

nk=0

k,nbk,n(x), (1)

where eachk,n ,k = 0, 1, . . . , n, is a real number and

bk,n(x)=n

k

x

k

(1x)nk

. (2)

The coefficientsk,n are calledBernstein coefficients and the polynomials b0,n(x), b1,n(x) , . . . , bn,n(x)are calledBernstein basis polynomialsof degreen. Define thenth Bernstein polynomial forfto be

Bn(f;x)=

nk=0

f

k

n

bk,n(x).

In 1912, Bernstein showed the following result[2,7]:

The Bernstein Theorem. Let f be a continuous function defined on the closed interval[0, 1]. For any >0, there exists a positive integer M such that for all x in[0, 1]and integer m M , we have

|f(x)Bm(f;x)| < .

Note that the functionBm(f;x)is a polynomial onx. Thus, on the basis of the Bernstein Theorem,the Weierstrass Approximation Theorem holds. Given a power-form polynomial gof degreen, it iswell known that for any m n,gcan be uniquely converted into a Bernstein polynomial of degreem[5]. Combining this fact with the Bernstein Theorem, we have the following corollary.

Corollary 1. Let g be a polynomial of degree n. For any >0, there exists a positive integer Mn suchthat for all x in[0, 1]and integer m M , we have

m

k=0

k,m g

km

bk,m(x) < ,

where0,m, 1,m, . . . , m,msatisfy g(x)=m

k=0k,mbk,m(x).

In the first part of the paper, we prove a stronger result than this:

Theorem 1. Let g be a polynomial of degree n 0. For any >0, there exists a positive integer M nsuch that for all integers m M and k= 0, 1, . . . , m, we have

k,m g

k

m

< ,

where0,m, 1,m, . . . , m,msatisfy g(x)=m

k=0k,mbk,m(x).

(CombiningTheorem 1with the fact thatm

k=0bk,m(x)= 1, we can easily proveCorollary 1.)In the second part of the paper, we consider a subset of Bernstein polynomials: those with

coefficients that are all in the unit interval[0, 1].


3/16

450 W. Qian et al. / European Journal of Combinatorics 32 (2011) 448463

Definition 1. DefineUto be the set of Bernstein polynomials with coefficients that are all in the unitinterval[0, 1]:

U = p(x)| n 1, 0 0,n, 1,n, . . . , n,n 1, such thatp(x)=n

k=0

k,nbk,n(x) .The question that we ask is: which polynomials can be converted into Bernstein polynomials inU?

Definition 2. Define the setVto be the set of polynomials which are either identically equal to 0 orequal to 1, or map the open interval(0, 1)into(0, 1)and the points 0 and 1 into the closed interval[0, 1], i.e.,

V = {p(x)|p(x) 0, orp(x) 1, or 0 < p(x)


4/16


The remainder of the paper is organized as follows. In Section2,we present some mathematicalpreliminaries pertaining to Bernstein polynomials. In Section3,we proveTheorem 1.On the basis ofthis theorem, in Section4,we proveTheorem 2.Finally, we conclude the paper with a discussion onapplications of these theorems to our research in probabilistic computation with digital circuits.

2. Properties of Bernstein polynomials

We list some pertinent properties of Bernstein polynomials.

(a) Thepositivityproperty:For allk = 0, 1, . . . , nand allxin[0, 1], we have

bk,n(x) 0. (3)

(b) Thepartition of unityproperty:The binomial expansion of the left-hand side of the equality (x+ (1 x))n = 1 shows that thesum of all Bernstein basis polynomials of degree nis the constant 1, i.e.,

nk=0

bk,n(x)= 1. (4)

(c) Converting power-form coefficients to Bernstein coefficients:The set of Bernstein basis polynomials b0,n(x), b1,n(x) , . . . , bn,n(x) forms a basis of the vector spaceof polynomials of real coefficients and degree no more than n [5]. Each power basis functionxj canbe uniquely expressed as a linear combination of then+1 Bernstein basis polynomials:

xj =

nk=0

jkbk,n(x), (5)

forj = 0, 1, . . . , n. To determine the elements of the transformation matrix, we write

xj =xj(x+(1x))nj

and perform a binomial expansion on the right-hand side. This gives

xj =

nk=j

k

j

n

j

bk,n(x),forj = 0, 1, . . . , n. Therefore, we have

jk =

jk =

k

j

n

j

1, forj k

0, forj > k.

(6)

Suppose that a power-form polynomial of degree no more thannis

g(x)=

nk=0

ak,nxk (7)

and the Bernstein polynomial of degreenofgis

g(x)=

nk=0

k,nbk,n(x). (8)

Substituting Eqs.(5)and(6)into Eq.(7)and comparing the Bernstein coefficients, we have

k,n =

nj=0

aj,njk =

kj=0

k

j

n

j

1aj,n. (9)

Eq.(9)provides a means for obtaining Bernstein coefficients from power-form coefficients.


5/16


(d) Degree elevation:

On the basis of Eq.(2),we have that for all k = 0, 1, . . . , m,

1

m+1k

bk,m+1(x)+1

m+1k+1

bk+1,m+1(x) = x

k(1x)m+1k +xk+1(1x)mk

= xk(1x)mk =1m

k

bk,m(x),or

bk,m(x) =

m

k

m+1

k

bk,m+1(x)+

m

k

m+1

k+1

bk+1,m+1(x)=

m+1k

m+1bk,m+1(x)+

k+1

m+1bk+1,m+1(x). (10)

Given a power-form polynomialgof degreen, for anym n,gcan be uniquely converted into aBernstein polynomial of degree m. Suppose that the Bernstein polynomials of degree m and degree

m+1 ofgarem

k=0

k,mbk,m(x)andm+1k=0

k,m+1bk,m+1(x), respectively. We have

mk=0

k,mbk,m(x)=

m+1k=0

k,m+1bk,m+1(x). (11)

Substituting Eq.(10)into the left-hand side of Eq.(11)and comparing the Bernstein coefficients,we have

k,m+1 =

0,m

, fork = 0k

m+1k1,m+

1

k

m+1

k,m, for 1 k m

m,m, fork = m +1.

(12)

Eq. (12) provides a means for obtaining the coefficients of the Bernstein polynomial of degree m+1ofgfrom the coefficients of the Bernstein polynomial of degree m ofg. We will call this proceduredegree elevation.

For convenience, given a Bernstein polynomialg(x)=n

k=0k,nbk,n(x), we can also express it as

g(x)=

n

k=0

ck,nxk(1x)nk, (13)

where

ck,n =n

k

k,n, (14)

fork = 0, 1, . . . , n. Substituting Eq.(14)into Eq.(12),we have

ck,m+1 =

c0,m, fork = 0ck1,m+ ck,m, for 1 k mcm,m, fork = m +1.

(15)

3. A proof ofTheorem 1

Suppose that the polynomialgis of degreen. Applying Eq.(15)recursively, we can expressck,masa linear combination ofc0,n, c1,n, . . . , cn,n.


6/16


Lemma 1. Let g be a polynomial of degree n. For any m n, suppose that the Bernstein polynomial ofdegree m of g is g(x) =

mk=0ck,mx

k(1 x)mk. Let ck,m = 0for all k < 0and all k > m. Then for allk= 0, 1, . . . , m, we have

ck,m =

mni=0

mni

ckm+n+i,n. (16)

Proof. We prove the lemma by induction on mn.

Base case: Form n =0, the right-hand side of Eq. (16)reduces to

0

0

ck,n = ck,m, so the equation

holds.

Inductive step: Suppose that Eq.(16)holds for somem n and allk = 0, 1, . . . , m. Considerm+1.Since we assume thatc1,m =cm+1,m =0, Eq.(15)can be written as

ck,m+1 =ck1,m+ ck,m, (17)

for all k= 0, . . . , m + 1. With our convention that ci,n =0 for all i< 0 and i > n, it is easily seen that

c1,m =0 =

mni=0

mn

i

c1m+n+i,n, cm+1,m =0 =

mni=0

mn

i

cm+1m+n+i,n.

Combining this with the induction hypothesis, we conclude that for all k = 1, 0, . . . , m, m+1,

ck,m =

mni=0

mn

i

ckm+n+i,n. (18)

On the basis of Eqs.(17)and(18),for allk = 0, 1, . . . , m+1, we have

ck,m+1 =mni=0

mn

i

ck1m+n+i,n+

mnj=0

mn

j

ckm+n+j,n.

In the first sum, we change the summation index to j = i 1. We obtain

ck,m+1 =

mn1j=1

mn

j+1

ckm+n+j,n+

mnj=0

mn

j

ckm+n+j,n

=

mn

0

ckm+n1,n+

mn1

j=0[

mn

j+1

+

mn

j

]ckm+n+j,n+

mn

mn

ck,n.

Applying the basic formula

r

q

=

r1

q1

+

r1

q

, we obtain

ck,m+1 =ckm+n1,n+

mn1j=0

m+1n

j+1

ckm+n+j,n+ ck,n

=

m+1ni=0

m+1n

i

ckm1+n+i,n.

Thus Eq.(16)holds form+1. By induction, it holds for allm k.

Remark. Eq.(16)can be formulated as

ck,m =

min{k,n}i=max{0,km+n}

mn

ki

ci,n, (19)


7/16


for allm nandk = 0, 1, . . . , m. Indeed, in Eq.(16),first use the basic formula

r

q

=

r

rq

and

then change the summation index to j = k m+n+ito obtain

ck,m =

mn

i=0 mn

mni ckm+n+i,n =

k

j=km+n

mnkj

cj,n.Note thatcj,n =0 implies 0 j n. This yields Eq.(19).

Lemma 2. Let n be a positive integer. For all integer m, k and i such that

m > n, 0 k m, max{0, km+n} i min{k, n}, (20)

we have

k

m

i 1

k

m

ni

mn

ki

m

k

n2

m. (21)

Proof. For simplicity, we define =( km

)i(1 km

)ni

mnki

( mk)

. Nowmn

ki

m

k

= (mn)!(ki)!(mnk+i)!

k!(mk)!

m!

=k(k1) (ki+1)(mk)(mk1) (mnk+i+1)

m(m1) (mn+1)

=

i1

j=0

kj

mj

ni1

j=0

mkj

mij=

i1

j=0

1mk

mj ni1

j=0

1ki

mij . (22)We obtain an upper bound for

mnki

( mk)

by replacingjin Eq.(22)with its least value, 0:mn

ki

m

k

i1j=0

1

mk

m

ni1j=0

1

ki

mi

=

k

m

i mk

mi

ni.

We need the following simple inequality: for real numbers 0 x y 1 and a non-negativeintegerl,

y

l

x

l

=(yx)

l1j=0 y

j

x

l1j

(yx)l. (23)

From Eq.(20),we obtain 0 i min{k, n} k m and so we can use Eq.(23)for

0 x =mk

m

mk

mi=y 1, l = n i 0.

We obtain

=

k

m

i 1

k

m

ni

mn

ki

m

k

k

m

i mk

m

ni

mk

mi

ni

=

km

i mkmi

ni

mkm

ni

k

m

i mk

mi

mk

m

(ni)=

k

m

i(mk)i(ni)

(mi)m.


8/16


Since 0 km

1, 0 mkmi

1, and 0 i n, we obtain

k

m

i(mk)i(ni)

(mi)m

i(ni)

m>

n2

m.

Therefore,

=

k

m

i 1

k

m

ni

mn

ki

m

k

> n2m

. (24)

Similarly, we obtain a lower bound for

mnki

( mk)

by replacing the indexj in Eq.(22)withi in the first

product and withniin the second product, obtaining

mn

ki mk =

i1

j=0

1mk

mj ni1

j=0

1ki

mij

i1j=0

1

mk

mi

ni1j=0

1

ki

mn

=

k i

mi

i mnk+i

mn

ni

k i

mi

i mnk+i

mn+i

ni.

Thus, proceeding as above, we have

= k

mi

1k

mni

mn

ki

mk

k

mi

mk

m ni

k i

mii

mnk+i

mn+i ni

=

k

m

i

k i

mi

imk

m

ni

+

mk

m

ni

mnk+i

mn+i

nik i

mi

i.

Due to Eq.(20),we have

0ki

mi

k

m1, 0

mnk+i

mn+i

mk

m1,

and so we obtain

k

m

i

k i

mi

i+

mk

m

ni

mnk+i

mn+i

ni. (25)

Applying Eq.(23)twice to the right-hand side of Eq.(25),we obtain

i

k

m

ki

mi

+(ni)

mk

m

mnk+i

mn+i

=i2

m

mk

mi

+(ni)2

m

k

mn+i

.

From Eq.(20),we have

0mk

mi1, 0

k

mn+i1.


9/16


Therefore,

=

k

m

i 1

k

m

ni

mn

ki

m

k

i2 +(ni)2

m

ni+n(ni)

m=

n2

m. (26)

Eqs.(24)and(26)together yield Eq.(21). Now we give a proof ofTheorem 1.

Theorem 1. Let g be a polynomial of degree n 0. For any >0, there exists a positive integer M nsuch that for all integer m M and k= 0, 1, . . . , m, we havek,m g

k

m

< ,where0,m, 1,m, . . . , m,msatisfy that g(x)=

mk=0k,mbk,m(x).

Proof. Forn = 0,g is a constant polynomial. Suppose that g(x) = y, wherey is a constant value.We select M = 1. Then, for all integers m Mand all integers k = 0, 1, . . . , m, we havek,m =

y = g

km

. Thus, the theorem holds.

For n > 0, we select Msuch that M > max{ n2

ni=0|ci,n|, 2n}, where the real numbers c0,n,

c1,n, . . . , cn,nsatisfy

g(x)=

ni=0

ci,nxi(1x)ni. (27)

Now consider anym M. Since

2n max

n2

n

i=0|ci,n|, 2n

< M m,

we havemn > n. Consider the following three cases fork.

1. The case where n k m n. Here max{0, k m + n} = 0 and min{k, n} = n. Thus, thesummation indices in Eq.(19)range from 0 ton. Therefore,

k,m =ck,m

m

k

= ni=0

mn

ki

m

k

ci,n. (28)Substitutingxwith k

min Eq.(27),we have

g

k

m =

n

i=0ci,n

k

mi

1

k

mni

. (29)

ByLemma 2,since 0< n < m and 0 k m, Eq.(21)holds for all 0= max{0, k m + n} i min{k, n} = n. Thus, by Eqs.(21),(28)and(29)and the well-known inequality|

xi|

|xi|,

we havek,m g

k

m

=

ni=0

mn

ki

m

k

km

i 1

k

m

nici,n

ni=0

mn

ki

m

k

km

i 1

k

m

ni |ci,n| n2

m

ni=0

|ci,n|.

Since n2

ni=0|ci,n|< M m, we haven2

m

ni=0

|ci,n|< . (30)

Therefore, for alln k m n, we have|k,m g(k

m)|< .


10/16


2. The case where 0 k < n. Sincem > 2n, we havek m + n < k n < 0. Thus, max{0, km+n} =0 and min{k, n} =k. Thus, the summation indices in Eq. (19) range from0 to k. Therefore,

k,m =ck,m

mk =

k

i=0

mn

ki mk ci,n. (31)

Whenk+1 i n, we have that 1 k +1 i and so

k

m

i 1

k

m

ni =

k

m

k

m

i1 1

k

m

ni km < nm n2

m. (32)

ByLemma 2,since 0< n < m and 0 k m, Eq.(21)holds for all 0= max{0, k m + n} i min{k, n} =k. Thus, by Eqs.(21)and(29)(32)and the inequality|

xi|

|xi|, we have

k,m gk

m = k

i=0

mn

ki

mk ci,n

n

i=0

k

mi

1k

mni

ci,n=

k

i=0

mn

ki

m

k

km

i 1

k

m

nici,n

ni=k+1

k

m

i 1

k

m

nici,n

ki=0

mn

ki

m

k

km

i 1

k

m

ni |ci,n| +n

i=k+1

k

m

i 1

k

m

ni |ci,n|

n2

m

ni=0

|ci,n|< .

3. The case where m n< k m. Since m> 2n, we have n< m n< k. Thus, max{0, k m + n} =km +nand min{k, n} =n. Now, the summation indices in Eq.(19)range fromkm +nton.Therefore,

k,m =ck,m

m

k

= ni=km+n

mn

ki

m

k

ci,n. (33)When 0 i k m +n1, we have that 1 m +1k n i. Thus,

k

m

i 1

k

m

ni

=

1

k

m

k

m

i 1

k

m

ni1

mk

m 0. By contraposition, suppose that h(0) 0. Since h(0) = 0, we have

h(0) 0.

Sinceg(x) > 0 for allxin(0, 1), we haveh(x) = g(x)

xr(1x)s > 0 for allx in(0, 1). In view of the fact

that h(0) >0 and h(1) >0, we have h(x) >0, for allx in [0, 1]. Since h(x) is continuous on the closedinterval[0, 1], it attains its minimum valuemhon[0, 1]. Clearly,mh >0.

Let2 = mh > 0. ByTheorem 1,there exists a positive integer M2 n rs, such that for allintegersd M2 and k = 0, 1, . . . , d, we have|k,d h(

k

d)| < 2, where0,d, 1,d, . . . , d,d satisfy

that

h(x)=dk=0

k,dbk,d(x). (37)

Thus, for alld M2and allk = 0, 1, . . . , d,

k,d >h

k

d

2 mh mh =0.

Combining Eqs.(36)and(37),we have

g(x) = xr(1x)sh(x)= xr(1x)sd

k=0 k,dbk,d(x)= xr(1x)s

dk=0

k,d

d

k

xk(1x)dk

=

dk=0

k,d

d

k

d+r+s

k+r

d+r+sk+r

xk+r(1x)d+sk =

d+rk=r

kr,d

d

kr

d+r+s

k

bk,d+r+s(x)

=

d+r+sk=0

k,d+r+sbk,d+r+s(x),

wherek,d+r+sare the coefficients of the Bernstein polynomial of degreed+r+sofgand

k,d+r+s =

0, for 0 k < rand d+r 0, forrk d +r.Thus, whenm = d +r+s M2+ r+s, we have for allk = 0, 1, . . . , m,

k,m 0. (38)

According to Eqs.(35)and(38),if we select an m0 max{M1, M2 + r + s}, then g(x) can beexpressed as a Bernstein polynomial of degree m0:

g(x)=

m0k=0

k,m0 bk,m0 (x),

with 0 k,m0 1, for allk = 0, 1, . . . , m0. Therefore,g U.


13/16


Theorem 5. Let g be a polynomial of degree n mapping the open interval (0, 1)into(0, 1)with g(0)= 0and g(1)= 1. Then gU .

Proof. Denote byrthe multiplicity of 0 as a root ofg(x). We can factorizeg(x)as

g(x)= xr

h(x), (39)where h(x) is a polynomial satisfying h(0) =0. By a reasoning similar to that in the proof ofTheorem 4,

we obtainh(0) > 0. Since for all x in (0, 1],h(x) = g(x)

xr > 0, we have for all x in [0, 1],h(x) > 0.

Since h(x) is continuous on the closed interval [0, 1], it attains its minimum value mhon [0, 1]. Clearly,mh >0.

Let 1 =mh >0. By Theorem 1, there exists a positive integer M1 n rsuch that for all integersd M1andk = 0, 1, . . . , d, we have|k,d h(

k

d)|< 1, where0,d, 1,d, . . . , d,dsatisfy

h(x)=

d

k=0k,dbk,d(x). (40)

Thus, for alld M1and allk = 0, 1, . . . , d,

k,d >h

k

d

1 mh mh =0.

Combining Eqs.(39)and(40),we have

g(x) = xrh(x)= xrd

k=0

k,dbk,d(x)= xr

dk=0

k,d

d

k

xk(1x)dk

=

d

k=0

k,d d

kd+rk+r

d+rk+r

xk+r(1x)dk =d+rk=r

kr,d d

krd+r

k

bk,d+r(x)

=

d+rk=0

k,d+rbk,d+r(x),

wherek,d+rare the coefficients of the Bernstein polynomial of degree d+rofgand

k,d+r =

0, for 0 k < r

kr,d

d

kr

d+rk >0, forr k d +r.

Thus, whenm = d +r M1+ r, we have for all k = 0, 1, . . . , m,

k,m 0. (41)

Let

g =1 g(x). (42)

Then g maps the open interval (0, 1) into (0, 1) with g(0) = 1, g(1) = 0. Denote by s themultiplicity of 1 as a root ofg(x). Thus, we can factorizeg(x)as

g(x)= (1x)sh(x), (43)

where h(x) is a polynomial satisfying that h(1) = 0. As in the proof ofTheorem 4, we obtain

h(1) > 0. Since for allx in [0, 1),h(x) = g(x)

(1x)s > 0, we have for allx [0, 1],h(x) > 0. Since

h(x) is continuous on the closed interval [0, 1], it attains its minimum value mh on[0, 1]. Clearly,mh >0.


14/16


Let 2 =mh >0. By Theorem 1, there exists a positive integer M2 n s such that for all integers

q M2andk = 0, 1, . . . , q, we have|

k,q h( k

q)|< 2, where

0,q,

1,q, . . . ,

q,qsatisfy

h(x)=

q

k=0

k,qbk,q(x). (44)

Thus, for allq M2and allk = 0, 1, . . . , q,

k,q >h

k

q

2 m

h m

h =0.

Combining Eqs.(42)(44),we have

g(x) = 1g(x)= 1 (1x)sh(x)= 1 (1x)sq

k=0

k,qbk,q(x)

= 1(1x)s

qk=0

k,qq

k

xk

(1x)qk

=1

qk=0

k,q qk q+s

k

q+sk

xk

(1x)q+sk

.

Further using Eq.(4),we obtain

g(x)=

q+sk=0

bk,q+s(x)

qk=0

k,q

q

k

q+s

k

bk,q+s(x)= q+sk=0

k,q+sbk,q+s(x),

where thek,q+ss are the coefficients of the Bernstein polynomial of degreeq+sofg:

k,q+s =

1k,q

q

k

q+sk


15/16


Proof. Let polynomial h = 1 g. Then h maps (0, 1) into (0, 1) with 0 h(0), h(1) < 1. ByTheorem 4,h U. ByLemma 3,g =1 his also in the setU.

Corollary 3. Let g be a polynomial of degree n mapping the open interval (0, 1) into (0, 1) with g(0)= 1and g(1)= 0. Then gU .

Proof. Let the polynomial h = 1 g. Thenh maps (0, 1)into(0, 1)withh(0) = 0, h(1) = 1. ByTheorem 5,h U. ByLemma 3,g =1 his also in the setU.

CombiningTheorems 4and5,Corollaries 2and3,we show thatV U.

Theorem 6.

V U.

Proof. On the basis of the definition ofV, for any polynomialp V, we have one of the following fivecases.

1. The case wherep 0 orp 1. In the proof ofTheorem 3,we have shown that 0 Uand 1 U.Thusp

U.

2. The case wherepmaps the open interval(0, 1)into(0, 1)with 0 p(0),p(1)


16/16


The Boolean functionf evaluates to 1 if and only if the 3-tuple (x1,x2,x3) takes values from the set{(0, 0, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1)}. The probability of the output being 1 is

po = Pr(f =1) = Pr(x1,x2,x3 : (x1,x2,x3) {(0, 0, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1)})

= Pr(x1 =0,x2 =0,x3 =1)+Pr(x1 =0,x2 =1,x3 =1)

+ Pr(x1 =1,x2 =1,x3 =0)+Pr(x1 =1,x2 =1,x3 =1).

Ifx1,x2, andx3are independent random Boolean variables with probability p1,p2, andp3 of being 1,respectively, then we obtain

po = (1p1)(1p2)p3+ (1p1)p2p3+ p1p2(1p3)+p1p2p3

= (1p1)p3+ p1p2

= p1p2+ p3 p1p3, (46)

which confirms that the function computed by stochastic logic is a multivariate polynomial onargumentsp1,p2,andp3 with integer coefficients and with the degree of each variable no more than 1.

In design problems, we encounter univariate polynomials that have real coefficients and degreegreater than 1. Sometimes it is possible to implement these by setting some of the probabilities pito be a common variablex and the others to be constants. For example, if we set p1 = p3 = xand

p2 = 0.75 in Eq.(46),then we obtain the polynomial g(x) = 1.75x x2. With different underlying

Boolean functions and different assignments of probability values, we can implement many differentunivariate polynomials.

An interesting and yet practical question is: which univariate polynomials can be implemented bystochastic logic? Define the setWto be the set of (univariate) polynomials that can be implemented.We are interested in characterizing the setW.

In[9]we showed that U W, i.e., if a polynomial can be expressed as a Bernstein polynomialwith all coefficients in the unit interval, then the polynomial can be implemented by stochastic logic.In this paper, we proved thatV =U. Thus, we haveV W.

Further, in [9] we showed that W V, i.e., if a polynomial can be implemented by stochastic logic,then it is either identically equal to 0 or equal to 1, or it maps the open interval(0, 1)into the openinterval (0, 1) and maps the points 0 and 1 into the closed interval [0, 1]. Therefore, we conclude thatW = V, i.e., a polynomial can be implemented by stochastic logic if and only if it is either identicallyequal to 0 or equal to 1, or it maps the open interval (0, 1)into the open interval(0, 1)and maps thepoints 0 and 1 into the closed interval[0, 1].

This necessary and sufficient conditions allows us to answer the question of whether any givenpolynomial can be implemented by stochastic logic. On the basis of the mathematics, we haveproposed a constructive design method [9]. An overview of the method and its applications in circuitdesign will appear in a forthcoming Research Highlights article in Communications of the ACM [10].

References

[1] J. Berchtold, A. Bowyer, Robust arithmetic for multivariate Bernstein-form polynomials, Computer-Aided Design 32 (11)(2000) 681689.

[2] S.N. Bernstein, Dmonstration du thorme de Weierstrass fonde sur le calcul des probabilits, Communications of theKharkov Mathematical Society 13 (1912) 12.

[3] B. Brown,H. Card, Stochastic neural computation I: computationalelements,IEEE Transactionson Computers50 (9)(2001)891905.

[4] P.J. Davis, Interpolation and Approximation, Blaisdell Publishing Company, 1963.[5] R. Farouki, V. Rajan, On the numerical condition of polynomials in Bernstein form, Computer Aided Geometric Design 4

(3) (1987) 191216.[6] B. Gaines, Stochastic computing systems, in: Advances in Information Systems Science, vol. 2, Plenum, 1969, pp. 37172.

ch. 2.[7] K.M. Levasseur, A probabilistic proof of the Weierstrass approximation theorem, The American Mathematical Monthly 91

(4) (1984) 249250.

[8] G. Lorentz, Bernstein Polynomials, University of Toronto Press, 1953.[9] W. Qian, M.D. Riedel, The synthesis of robust polynomial arithmetic with stochastic logic, in: Design Automation

Conference, 2008, pp. 648653.[10] W. Qian, M.D. Riedel, Synthesizing logical computation on stochastic bit streams, Communications of the ACM, 2010

(submitted for publication).

Documents

Approximation and Bernstein Polynomials