In this section, we use the generating functions for the Bernstein basis functions to derive a family of functional equations. Using these equations, we derive a collection of identities for the Bernstein basis functions.

### 3.1 Sums and alternating sums

From (3), we get the following functional equations:

\sum _{k=0}^{\mathrm{\infty}}{f}_{\mathbb{B},k}(x,t)={e}^{t}

(8)

and

\sum _{k=0}^{\mathrm{\infty}}{(-1)}^{k}{f}_{\mathbb{B},k}(x,t)={e}^{(1-2x)t}.

(9)

**Theorem 3.1** (Sum of the Bernstein basis functions)

\sum _{k=0}^{n}{B}_{k}^{n}(x)=1.

*Proof* From (8), one finds that

\sum _{k=0}^{\mathrm{\infty}}{f}_{\mathbb{B},k}(x,t)=\sum _{n=0}^{\mathrm{\infty}}\frac{{t}^{n}}{n!}.

(10)

By combining (2) and (10), we get

\sum _{n=0}^{\mathrm{\infty}}(\sum _{k=0}^{n}{B}_{k}^{n}(x))\frac{{t}^{n}}{n!}=\sum _{n=0}^{\mathrm{\infty}}\frac{{t}^{n}}{n!}.

Comparing the coefficients of \frac{{t}^{n}}{n!} on both sides of the above equation, we arrive at the desired result. □

**Theorem 3.2** (Alternating sum of the Bernstein basis functions)

\sum _{k=0}^{n}{(-1)}^{k}{B}_{k}^{n}(x)={(1-2x)}^{n}.

*Proof* By combining (9) and (10), we obtain

\sum _{n=0}^{\mathrm{\infty}}(\sum _{k=0}^{n}{(-1)}^{k}{B}_{k}^{n}(x))\frac{{t}^{n}}{n!}=\sum _{n=0}^{\mathrm{\infty}}\frac{{(1-2x)}^{n}{t}^{n}}{n!}.

Comparing the coefficients of \frac{{t}^{n}}{n!} on both sides of the above equation, we arrive at the desired result. □

**Remark 3.3** Goldman [6], [[5], Chapter 5, pp.299-306] derived the formula for the alternating sum of the Bernstein basis functions algebraically.

### 3.2 Subdivision

From (3), we have the following functional equation:

{f}_{\mathbb{B},j}(xy,t)={f}_{\mathbb{B},j}(x,ty){e}^{t(1-y)}.

(11)

From this functional equation, we get the following identity which is the basis for subdivision of Bezier curves, *cf.* [4–6, 15].

**Theorem 3.4**

{B}_{j}^{n}(xy)=\sum _{k=j}^{n}{B}_{j}^{k}(x){B}_{k}^{n}(y).

*Proof* By equations (3) and (11),

\sum _{n=j}^{\mathrm{\infty}}{B}_{j}^{n}(xy)\frac{{t}^{n}}{n!}=(\sum _{n=0}^{\mathrm{\infty}}{B}_{j}^{n}(x){y}^{n}\frac{{t}^{n}}{n!})\left(\sum _{n=0}^{\mathrm{\infty}}\frac{{(1-y)}^{n}{t}^{n}}{n!}\right).

Therefore

\sum _{n=j}^{\mathrm{\infty}}{B}_{j}^{n}(xy)\frac{{t}^{n}}{n!}=\sum _{n=j}^{\mathrm{\infty}}(\sum _{k=j}^{n}{B}_{j}^{k}(x)\frac{{y}^{k}{(1-y)}^{n-k}}{k!(n-k)!}){t}^{n}.

Substituting (1) into the above equation, we arrive at the desired result. □

**Remark 3.5** Theorem 3.4 is a bit tricky to prove with algebraic manipulations. Goldman [6], [[5], Chapter 5, pp.299-306] proved this identity algebraically. He also proved the following related subdivision identities:

{B}_{j}^{n}((1-y)x+y)=\sum _{k=0}^{j}{B}_{j-k}^{n-k}(x){B}_{k}^{n}(y)

and

{B}_{j}^{n}((1-y)x+yz)=\sum _{k=0}^{n}(\sum _{p+q=j}{B}_{p}^{n-k}(x){B}_{q}^{k}(z)){B}_{k}^{n}(y).

For additional identities, see [6], [[5], Chapter 5, pp.299-306].

### 3.3 Formula for the monomials in terms of the Bernstein basis functions

Multiplying both sides of (3) by \left(\begin{array}{c}k\\ l\end{array}\right), we get

\left(\begin{array}{c}k\\ l\end{array}\right)\frac{{(xt)}^{k}}{k!}{e}^{t(1-x)}=\left(\begin{array}{c}k\\ l\end{array}\right)\sum _{n=0}^{\mathrm{\infty}}{B}_{k}^{n}(x)\frac{{t}^{n}}{n!}.

Summing both sides of the above equation over *k*, we obtain the following functional equation, which is used to derive a formula for the monomials in terms of the Bernstein basis functions:

\frac{{x}^{l}{t}^{l}}{l!}{e}^{t}=\sum _{k=0}^{\mathrm{\infty}}\left(\begin{array}{c}k\\ l\end{array}\right){f}_{\mathbb{B},k}(x,t).

(12)

**Theorem 3.6**

\left(\begin{array}{c}n\\ l\end{array}\right){x}^{l}=\sum _{l=0}^{k}\left(\begin{array}{c}k\\ l\end{array}\right){B}_{k}^{n}(x).

*Proof* Combining (2) and (12), we get

\frac{{x}^{l}}{l!}\sum _{n=0}^{\mathrm{\infty}}\frac{{t}^{n+l}}{n!}=\sum _{n=0}^{\mathrm{\infty}}(\sum _{k=0}^{n}\left(\begin{array}{c}k\\ l\end{array}\right){B}_{k}^{n}(x))\frac{{t}^{n}}{n!}.

Therefore

\sum _{n=0}^{\mathrm{\infty}}\left(\left(\begin{array}{c}n\\ l\end{array}\right){x}^{l}\right)\frac{{t}^{n}}{n!}=\sum _{n=0}^{\mathrm{\infty}}(\sum _{k=0}^{n}\left(\begin{array}{c}k\\ l\end{array}\right){B}_{k}^{n}(x))\frac{{t}^{n}}{n!}.

Comparing the coefficients of \frac{{t}^{n}}{n!} on both sides of the above equation, we arrive at the desired result. □

### 3.4 Differentiating the Bernstein basis functions

In this section we give higher-order derivatives of the Bernstein basis functions. We begin by observing that

{f}_{\mathbb{B},k}(x,t)={g}_{k}(t,x)h(t,x),

(13)

where

{g}_{k}(t,x)=\frac{{t}^{k}{x}^{k}}{k!}

and

Using Leibnitz’s formula for the *l* th derivative, with respect to *x*, we obtain the following *higher-order partial differential equation*:

\frac{{\partial}^{l}{f}_{\mathbb{B},k}(x,t)}{\partial {x}^{l}}=\sum _{j=0}^{l}\left(\begin{array}{c}l\\ j\end{array}\right)\left(\frac{{\partial}^{j}{g}_{k}(t,x)}{\partial {x}^{j}}\right)\left(\frac{{\partial}^{l-j}h(t,x)}{\partial {x}^{l-j}}\right).

(14)

From this equation, we arrive at the following theorem.

**Theorem 3.7**

\frac{{\partial}^{l}{f}_{\mathbb{B},k}(x,t)}{\partial {x}^{l}}=\sum _{j=0}^{l}\left(\begin{array}{c}l\\ j\end{array}\right){(-1)}^{l-j}{t}^{l}{f}_{\mathbb{B},k-j}(x,t).

(15)

*Proof* Formula (15) follows immediately from (14). □

Applying Theorem 3.7, we obtain a new derivation for the higher-order derivatives of the Bernstein basis functions.

**Theorem 3.8**

\frac{{d}^{l}{B}_{k}^{n}(x)}{d{x}^{l}}=\frac{n!}{(n-l)!}\sum _{j=0}^{l}{(-1)}^{l-j}\left(\begin{array}{c}l\\ j\end{array}\right){B}_{k-j}^{n-l}(x).

(16)

*Proof* By substituting the right-hand side of (2) into (15), we get

\sum _{n=0}^{\mathrm{\infty}}\left(\frac{{d}^{l}{B}_{k}^{n}(x)}{d{x}^{l}}\right)\frac{{t}^{n}}{n!}=\sum _{n=0}^{\mathrm{\infty}}(\sum _{j=0}^{l}{(-1)}^{l-j}\left(\begin{array}{c}l\\ j\end{array}\right){B}_{k-j}^{n}(x))\frac{{t}^{n+l}}{n!}.

Therefore

\sum _{n=0}^{\mathrm{\infty}}\left(\frac{{d}^{l}{B}_{k}^{n}(x)}{d{x}^{l}}\right)\frac{{t}^{n}}{n!}=\sum _{n=0}^{\mathrm{\infty}}(\sum _{j=0}^{l}{(-1)}^{l-j}\left(\begin{array}{c}l\\ j\end{array}\right)\left(\begin{array}{c}n\\ l\end{array}\right)l!{B}_{k-j}^{n-l}(x))\frac{{t}^{n}}{n!}.

Comparing the coefficients of \frac{{t}^{n}}{n!} on both sides of the above equation, we arrive at the desired result. □

Substituting l=1 into (16), we arrive at the following standard corollary.

**Corollary 3.9**

\frac{d}{dx}{B}_{k}^{n}(x)=n({B}_{k-1}^{n-1}(x)-{B}_{k}^{n-1}(x)),

*cf*. [1–16].

### 3.5 Recurrence relation

In the previous section we computed the derivative of (13) with respect to *x* to derive a derivative formula for the Bernstein basis functions. In this section we are going to differentiate (13) with respect to *t* to derive a recurrence relation for the Bernstein basis functions.

Using Leibnitz’s formula for the *v* th derivative, with respect to *t*, we obtain the following *higher-order partial differential equation*:

\frac{{\partial}^{v}{f}_{\mathbb{B},k}(x,t)}{\partial {t}^{v}}=\sum _{j=0}^{v}\left(\begin{array}{c}v\\ j\end{array}\right)\left(\frac{{\partial}^{j}{g}_{k}(t,x)}{\partial {t}^{j}}\right)\left(\frac{{\partial}^{v-j}h(t,x)}{\partial {t}^{v-j}}\right).

(17)

From the above equation, we have the following theorem.

**Theorem 3.10**

\frac{{\partial}^{v}{f}_{\mathbb{B},k}(x,t)}{\partial {t}^{v}}=\sum _{j=0}^{v}{B}_{j}^{v}(x){f}_{\mathbb{B},k-j}(x,t).

(18)

*Proof* Formula (18) follows immediately from (17). □

Using definitions (3) and (1) in Theorem 3.10, we obtain a recurrence relation for the Bernstein basis functions.

**Theorem 3.11**

{B}_{k}^{n}(x)=\sum _{j=0}^{v}{B}_{j}^{v}(x){B}_{k-j}^{n-v}(x).

(19)

*Proof* By substituting the right-hand side of (2) into (18), we get

\frac{{\partial}^{v}}{\partial {t}^{v}}(\sum _{n=0}^{\mathrm{\infty}}{B}_{k}^{n}(x)\frac{{t}^{n}}{n!})=\sum _{n=0}^{\mathrm{\infty}}(\sum _{j=0}^{v}{B}_{j}^{v}(x){B}_{k-j}^{n}(x))\frac{{t}^{n}}{n!}.

Therefore

\sum _{n=v}^{\mathrm{\infty}}{B}_{k}^{n}(x)\frac{{t}^{n-v}}{(n-v)!}=\sum _{n=0}^{\mathrm{\infty}}(\sum _{j=0}^{v}{B}_{j}^{v}(x){B}_{k-j}^{n}(x))\frac{{t}^{n}}{n!}.

From the above equation, we get

\sum _{n=v}^{\mathrm{\infty}}{B}_{k}^{n}(x)\frac{{t}^{n-v}}{(n-v)!}=\sum _{n=v}^{\mathrm{\infty}}(\sum _{j=0}^{v}{B}_{j}^{v}(x){B}_{k-j}^{n-v}(x))\frac{{t}^{n-v}}{(n-v)!}.

Comparing the coefficients of \frac{{t}^{n}}{n!} on both sides of the above equation, we arrive at the desired result. □

We also computed the derivative of (3) with respect to *t* to derive the following higher-order partial differential equation:

\frac{{\partial}^{v}{f}_{\mathbb{B},k}(x,t)}{\partial {t}^{v}}=\sum _{j=0}^{v}\left(\begin{array}{c}v\\ j\end{array}\right){B}_{v-j}^{v}(x){f}_{\mathbb{B},k-v+j}(x,t).

By using the above equation, we derive another recurrence relation for the Bernstein basis functions as follows:

{B}_{k}^{n}(x)=\sum _{j=0}^{v}\left(\begin{array}{c}v\\ j\end{array}\right){B}_{v-j}^{v}(x){B}_{k-v+j}^{n-v}(x).

**Remark 3.12** Setting v=1 in (19), one obtains the standard recurrence

{B}_{k}^{n}(x)=(1-x){B}_{k}^{n-1}(x)+x{B}_{k-1}^{n-1}(x).

### 3.6 Degree raising

In this section we present a functional equation which we apply to provide a new proof of the degree raising formula for the Bernstein polynomials.

From (3), we obtain the following *functional equation*:

{(xt)}^{d}{f}_{\mathbb{B},k}(x,t)=\frac{(k+d)!}{k!}{f}_{\mathbb{B},k+d}(x,t).

Therefore

{x}^{d}{B}_{k}^{n}(x)=\frac{n!(k+d)!}{k!(n+d)!}{B}_{k+d}^{n+d}(x).

(20)

Substituting d=1 into the above equation, we have

x{B}_{k}^{n}(x)=\frac{k+1}{n+1}{B}_{k+1}^{n+1}(x).

(21)

The above relation can also be proved by (1), *cf.* [4–6].

From (3), we also get the following functional equation:

{(xt)}^{-d}{f}_{\mathbb{B},k}(x,t)=\frac{(k-d)!}{k!}{f}_{\mathbb{B},k-d}(x,t).

Therefore

{(1-x)}^{d}{B}_{k}^{n}(x)=\frac{n!(n+d-k)!}{(n+d)!(n-k)!}{B}_{k}^{n+d}(x).

Substituting d=1, we have

(1-x){B}_{k}^{n}(x)=\frac{(n+1-k)}{(n+1)}{B}_{k}^{n+1}(x).

(22)

Adding (21) and (22), we get the standard degree elevation formula

{B}_{k}^{n}(x)=\frac{1}{n+1}((k+1){B}_{k+1}^{n+1}(x)+(n+1-k){B}_{k}^{n+1}(x)).