**Proof.**

(i) If we put 1 instead of *x* in the generating function of the (*p*, *q*)-cosine Bernoulli polynomials, we find the following:

$$\begin{split} \sum\_{n=0}^{\infty} \\_ {\mathbb{C}} B\_{n,p,q}(1,y) \frac{t^n}{[n]\_{p,q}!} &= \frac{t}{\varepsilon\_{p,q}(t)-1} \left( \varepsilon\_{p,q}(t) - 1 + 1 \right) \text{COS}\_{p,q}(ty) \\ &= t \text{COS}\_{p,q}(ty) + \frac{t}{\varepsilon\_{p,q}(t)-1} \text{COS}\_{p,q}(ty) . \end{split} \tag{40}$$

Using a property of the (*p*, *q*)-exponential function, *ep*,*<sup>q</sup>*(*x*)*Ep*,*<sup>q</sup>*(−*<sup>x</sup>*) = 1, in Equation (40), we obtain the following:

$$\begin{split} &\sum\_{n=0}^{\infty} cB\_{n,p,q}(1,y)\frac{t^n}{[n]\_{p,q}!} \\ &= \left(tc\_{p,q}(\text{tx})\text{CO}S\_{p,q}(\text{ty}) + \frac{t}{\varepsilon\_{p,q}(t)-1}e\_{p,q}(\text{tx})\text{CO}S\_{p,q}(\text{ty})\right)E\_{p,q}(-t\mathbf{x}) \\ &= \sum\_{n=0}^{\infty} \left( [n]\_{p,q}\text{C}\_{n-1,p,q}(\text{x},y) + \_{\subset}B\_{n,p,q}(\text{x},y) \right) \frac{t^n}{[n]\_{p,q}!} \sum\_{n=0}^{\infty} q^{\binom{n}{2}}(-\mathbf{x})^n \frac{t^n}{[n]\_{p,q}!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^n \begin{bmatrix} n \\ k \end{bmatrix}\_{p,q}(-1)^{n-k} q^{\binom{n-k}{2}} ([k]\_{p,q}\text{C}\_{k-1,p,q}(\text{x},y) + \_{\subset}B\_{k,p,q}(\text{x},y)) x^{n-k} \right) \frac{t^n}{[n]\_{p,q}!}, \end{split} \tag{41}$$

and we immediately derive the results.

(ii) By applying a similar process for proving (*i*) to the (*p*, *q*)-sine Bernoulli polynomials, we find Theorem 6 (*ii*).

$$\square$$

**Corollary 4.** *Setting p* = 1 *in Theorem 6, the following holds:*

$$\begin{aligned} \mathbf{u}(i) & \ \_ {\subset} B\_{n, \emptyset}(1, y) = \sum\_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}\_q (-1)^{n-k} q^{\binom{n-k}{2}} \left( [k]\_q \mathbb{C}\_{k-1, \emptyset}(\mathbf{x}, y) + \_ {\subset} B\_{k, q}(\mathbf{x}, y) \right) \mathbf{x}^{n-k} \\\ \mathbf{u}(i) & \ \_ {\subset} B\_{n, q}(1, y) = \sum\_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}\_q (-1)^{n-k} q^{\binom{n-k}{2}} \left( [k]\_q \mathbb{S}\_{k-1, q}(\mathbf{x}, y) + \_ {\subset} B\_{k, q}(\mathbf{x}, y) \right) \mathbf{x}^{n-k} \end{aligned} \tag{42}$$

*where <sup>C</sup>Bn*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) *denotes the q-cosine Bernoulli polynomials and <sup>S</sup>Bn*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) *denotes the q-sine Bernoulli polynomials.*

**Corollary 5.** *Setting p* = 1 *and q* → 1 *in Theorem 6, the following holds:*

$$\begin{aligned} (i) \quad & \,\_\text{C}B\_{\text{fl}}(1,y) = \sum\_{k=0}^{n} \binom{n}{k} (-1)^{n-k} \left( k \text{C}\_{k-1}(\mathbf{x}, y) + \,\_\text{C}B\_{\text{k}}(\mathbf{x}, y) \right) \, \mathbf{x}^{n-k} \\ (ii) \quad & \,\_\text{S}B\_{\text{fl}}(1,y) = \sum\_{k=0}^{n} \binom{n}{k} (-1)^{n-k} \left( k \text{S}\_{k-1}(\mathbf{x}, y) + \,\_\text{S}B\_{\text{k}}(\mathbf{x}, y) \right) \, \mathbf{x}^{n-k}, \end{aligned} \tag{43}$$

*where <sup>C</sup>Bn*(*<sup>x</sup>*, *y*) *is the cosine Bernoulli polynomials and <sup>S</sup>Bn*(*<sup>x</sup>*, *y*) *is the sine Bernoulli polynomials.*

**Theorem 7.** *For a nonnegative integer k and* |*p*/*q*| < 1*, we investigate*

$$\begin{split} B\_{n,p,q}(\mathbf{x}) &= \sum\_{k=0}^{\left[\frac{n}{2}\right]} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q} (-1)^k p^{(2k-1)k} y^{2k} \subset B\_{n-k,p,q}(\mathbf{x}, \mathbf{y}) \\ &+ \sum\_{k=0}^{\left[\frac{n-1}{2}\right]} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_{p,q} (-1)^k p^{(2k+1)k} y^{2k+1} \,^s B\_{n-(2k+1),p,q}(\mathbf{x}, \mathbf{y}), \end{split} \tag{44}$$

*where Bn*,*p*,*<sup>q</sup>*(*x*) *is the* (*p*, *q*)*-Bernoulli polynomials and* [*x*] *is the greatest integer not exceeding x.* *Symmetry* **2020**, *12*, 885

**Proof.** In [9], we observe the power series of (*p*, *q*)-cosine and (*p*, *q*)-sine functions as follows:

$$\cos\_{p,q}(\mathbf{x}) = \sum\_{n=0}^{\infty} (-1)^n p^{(2n-1)n} \frac{\mathbf{x}^{2n}}{[2n]\_{p,q}!}, \quad \sin\_{p,q}(\mathbf{x}) = \sum\_{n=0}^{\infty} (-1)^n p^{(2n+1)n} \frac{\mathbf{x}^{2n+1}}{[2n+1]\_{p,q}!}.\tag{45}$$

Let us consider (*p*, *q*)-cosine Bernoulli polynomials. If we multiply (*p*, *q*)-cosine Bernoulli polynomials and the (*p*, *q*)-cosine function to determine the relationship between (*p*, *q*)-Bernoulli polynomials and, combined (*p*, *q*)-cosine Bernoulli polynomials, and (*p*, *q*)-sine Bernoulli polynomials, we have

$$\sum\_{n=0}^{\infty} \, \_{\subset} B\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \cos\_{p,q}(ty) = \frac{t}{e\_{p,q}(t) - 1} e\_{p,q}(tx) \text{COS}\_{p,q}(ty) \cos\_{p,q}(ty). \tag{46}$$

The left-hand side of Equation (46) is transformed as

$$\begin{split} &\sum\_{n=0}^{\infty} \\_ {\mathbb{C}B}\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \cos\_{p,q}(ty) \\ &= \sum\_{n=0}^{\infty} \\_ {\mathbb{C}B}\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \sum\_{n=0}^{\infty} (-1)^n p^{(2n-1)n} y^{2n} \frac{t^{2n}}{[n]\_{p,q}!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^n \begin{bmatrix} n+k \\ 2k \end{bmatrix}\_{p,q}(-1)^k p^{(2k-1)k} y^{2k} \mathcal{C}\_{\mathbb{C}B} B\_{n-k,p,q}(x,y) \right) \frac{t^{n+k}}{[n+k]\_{p,q}!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\left[\frac{q}{2}\right]} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q}(-1)^k p^{(2k-1)k} y^{2k} \mathcal{C}\_{\mathbb{C}B} B\_{n-k,p,q}(x,y) \right) \frac{t^n}{[n]\_{p,q}!} . \end{split} \tag{47}$$

From Equations (46) and (47), we derive the following:

$$\begin{split} &\sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\left[\frac{n}{2}\right]} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q} (-1)^k p^{(2k-1)k} y^{2k} \mathbf{c}\_{\subset} \mathbf{B}\_{n-k,p,q}(\mathbf{x}, \mathbf{y}) \right) \frac{t^n}{[n]\_{p,q}!} \\ &= \frac{t}{\mathfrak{e}\_{p,q}(t) - 1} \mathbf{c}\_{p,q}(t\mathbf{x}) \mathbf{C} \mathbf{O} S\_{p,q}(\mathbf{t}\mathbf{y}) \mathbf{c}\_{p,q}(\mathbf{t}\mathbf{y}), \end{split} \tag{48}$$

where [*x*] is the greatest integer that does not exceed *x*.

From now on, let us consider the (*p*, *q*)-sine Bernoulli polynomials in a same manner of (*p*, *q*)-cosine Bernoulli polynomials. If we multiply *<sup>S</sup>Bn*,*p*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) and *sinp*,*<sup>q</sup>*(*ty*), we obtain

$$\sum\_{n=0}^{\infty} \,\_3B\_{n,p,q}(\mathbf{x},y) \frac{t^n}{[n]\_{p,q}!} \sin\_{p,q}(ty) = \frac{t}{\varepsilon\_{p,q}(t) - 1} \varepsilon\_{p,q}(t\mathbf{x}) S \text{IN}\_{p,q}(ty) \sin\_{p,q}(ty). \tag{49}$$

The left-hand side of Equation (49) can be changed as the following.

$$\begin{split} &\sum\_{n=0}^{\infty} \, \_SB\_{n,p,q}(\mathbf{x},y) \frac{t^n}{[n]\_{p,q}!} \sin\_{p,q}(ty) \\ &= \sum\_{n=0}^{\infty} \, \_SB\_{n,p,q}(\mathbf{x},y) \frac{t^n}{[n]\_{p,q}!} \sum\_{n=0}^{\infty} (-1)^n \, p^{(2n+1)n} y^{2n+1} \frac{t^{2n+1}}{[n]\_{p,q}!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\left[\frac{n-1}{2}\right]} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_{p,q}(-1)^k \, p^{(2k+1)k} y^{2k+1} \,\_SB\_{n-(2k+1),p,q}(\mathbf{x},y) \right) \frac{t^n}{[n]\_{p,q}!} .\end{split} \tag{50}$$

From Equations (49) and (50), we have the following:

$$\begin{aligned} &\sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{2k+1}\_{p,q} (-1)^k p^{(2k+1)k} y^{2k+1} \, \_\text{S}B\_{n-(2k+1),p,q}(x,y) \right) \frac{t^n}{[n]\_{p,q}!} \\ &= \frac{t}{\varepsilon\_{p,q}(t)-1} e\_{p,q}(tx) SIN\_{p,q}(ty) \sin\_{p,q}(ty), \end{aligned} \tag{51}$$

where [*x*] is the greatest integer that does not exceed *x*.

Here, we recall that

$$\left(\text{COS}\_{p,q}(\mathbf{x})\text{cos}\_{p,q}(\mathbf{x}) + \text{SIN}\_{p,q}(\mathbf{x})\text{sin}\_{p,q}(\mathbf{x})\right) = 1.\tag{52}$$

Using Equations (48) and (51) and applying the property of (*p*, *q*)-trigonometric functions, we find (*p*, *q*)-Bernoulli polynomials as follows:

$$\begin{split} &\sum\_{n=0}^{\infty} \left( \binom{\frac{n}{2}}{k-0} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q} (-1)^k p^{(2k-1)k} y^{2k} cB\_{n-k,p,q}(\mathbf{x},y) \right) \frac{t^n}{[n]\_{p,q}!} \\ &+ \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\left[ \frac{n-1}{2} \right]} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_{p,q} (-1)^k p^{(2k+1)k} y^{2k+1} s B\_{n-(2k+1),p,q}(\mathbf{x},y) \right) \frac{t^n}{[n]\_{p,q}!} \\ &= \frac{t}{c\_{p,q}(t)-1} e\_{p,q}(t\mathbf{x}) \left( \mathrm{CO}\_{p,q}(ty) \mathrm{cos}\_{p,q}(ty) + \mathrm{SIN}\_{p,q}(ty) \mathrm{sin}\_{p,q}(ty) \right) \\ &= \sum\_{n=0}^{\infty} B\_{n,p,q}(\mathbf{x}) \frac{t^n}{[n]\_{p,q}!}, \end{split} \tag{53}$$

where *Bn*,*p*,*<sup>q</sup>*(*x*) is the (*p*, *q*)-Bernoulli polynomials.

By comparing the coefficients of both sides of *t n*, we produce the desired result.

**Corollary 6.** *Setting p* = 1 *in Theorem 7, the following holds:*

$$\begin{split} B\_{n,q}(\mathbf{x}) &= \sum\_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}\_q (-1)^k y^{2k} \, \_\text{\subset} B\_{n-k,q}(\mathbf{x}, \mathbf{y}) \\ &+ \sum\_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{2k+1}\_q (-1)^k y^{2k+1} \, \_\text{\subset} B\_{n-(2k+1),q}(\mathbf{x}, \mathbf{y}), \end{split} \tag{54}$$

*where Bn*,*<sup>q</sup>*(*x*) *is the q-Bernoulli polynomials, <sup>C</sup>Bn*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) *denote the q-cosine Bernoulli polynomials, and <sup>S</sup>Bn*,*<sup>q</sup>*(*x*) *denote the q-sine Bernoulli polynomials.*

**Corollary 7.** *Setting y* = 1 *in Theorem 7, one holds:*

$$\begin{split} B\_{n,p,q}(\mathbf{x}) &= \sum\_{k=0}^{\lfloor \frac{n}{2} \rfloor} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q} (-1)^k p^{(2k-1)k} \,\_{\mathbb{C}}B\_{n-k,p,q}(\mathbf{x}, 1) \\ &+ \sum\_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_{p,q} (-1)^k p^{(2k+1)k} \,\_{\mathbb{S}}B\_{n-(2k+1),p,q}(\mathbf{x}, 1), \end{split} \tag{55}$$

*where Bn*,*p*,*<sup>q</sup>*(*x*) *is the* (*p*, *q*)*-Bernoulli polynomials and* [*x*] *is the greatest integers that does not exceed x.* *Symmetry* **2020**, *12*, 885

**Theorem 8.** *For a nonnegative integer k and* |*p*/*q*| < 1*, we derive*

$$\begin{aligned} &\sum\_{k=0}^{\left[\frac{n-1}{2}\right]} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_{p,q} (-1)^k p^{(2k+1)k} y^{2k+1} \mathcal{C}\_{\mathbb{C}} B\_{n-(2k+1),p,q}(x,y) \\ &= \sum\_{k=0}^{\left[\frac{n}{2}\right]} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q} (-1)^k p^{(2k-1)k} y^{2k} \mathcal{B}\_{\mathbb{C}} B\_{n-k,p,q}(x,y), \end{aligned} \tag{56}$$

*where* [*x*] *is the greatest integer not exceeding x.*

**Proof.** If we multiply *<sup>C</sup>Bn*,*p*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) and *sinp*,*<sup>q</sup>*(*ty*), then we find

$$\sum\_{n=0}^{\infty} \, \_CB\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \sin\_{p,q}(ty) = \frac{t}{\varepsilon\_{p,q}(t) - 1} \varepsilon\_{p,q}(tx) \text{COS}\_{p,q}(ty) \sin\_{p,q}(ty), \tag{57}$$

and the left-hand side of Equation (57) can be transformed as

$$\begin{split} &\sum\_{n=0}^{\infty} \, \_CB\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \sin\_{p,q}(ty) \\ &= \sum\_{n=0}^{\infty} \, \_CB\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \sum\_{n=0}^{\infty} (-1)^n p^{(2n+1)n} y^{2n+1} \frac{t^{2n+1}}{[2n+1]\_{p,q}!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\left[\frac{n-1}{2}\right]} \left[ \begin{matrix} n \\ 2k+1 \end{matrix} \right]\_{p,q} (-1)^k p^{(2k+1)k} y^{2k+1} \,\_CB\_{n-(2k+1),p,q}(x,y) \right) \frac{t^n}{[n]\_{p,q}!} .\end{split} \tag{58}$$

Similarly, we multiply the (*p*, *q*)-sine Bernoulli polynomials and (*p*, *q*)-cosine function as the follows:

$$\sum\_{n=0}^{\infty} \, \_S B\_{n,p,q}(x,y) \frac{t^n}{[n]\_{p,q}!} \cos\_{p,q}(ty) = \frac{t}{c\_{p,q}(t) - 1} \varepsilon\_{p,q}(tx) SIN\_{p,q}(ty) \cos\_{p,q}(ty). \tag{59}$$

The left-hand side of Equation (59) can be changed as

$$\begin{split} &\sum\_{n=0}^{\infty} \,\_3B\_{n,p,q}(\mathbf{x},\mathbf{y}) \frac{t^n}{[n]\_{p,q}!} \cos\_{p,q}(t\mathbf{y}) \\ &= \sum\_{n=0}^{\infty} \,\_3B\_{n,p,q}(\mathbf{x},\mathbf{y}) \frac{t^n}{[n]\_{p,q}!} \sum\_{n=0}^{\infty} (-1)^n p^{(2n-1)n} y^{2n} \frac{t^{2n}}{[2n]\_{p,q}!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{2k}\_{p,q} (-1)^k p^{(2k-1)k} y^{2k} \,\_3B\_{n-k,p,q}(\mathbf{x},\mathbf{y}) \right) \frac{t^n}{[n]\_{p,q}!} .\end{split} \tag{60}$$

In here, we recall that *sinp*,*<sup>q</sup>*(*x*)*COSp*,*<sup>q</sup>*(*x*) = *cosp*,*<sup>q</sup>*(*x*)*SINp*,*<sup>q</sup>*(*x*). From Equations (58) and (60), and the above property of (*p*, *q*)-trigonometric functions, we investigate

$$\sum\_{n=0}^{\infty} \binom{\left[\frac{n-1}{2}\right]}{k=0} \binom{n}{2k+1}\_{p,q} (-1)^k p^{(2k+1)k} y^{2k+1} \,\_\circ B\_{n-(2k+1),p,q}(x,y) \begin{cases} t^n\\ \left[n\right]\_{p,q} \end{cases}$$

$$= -\sum\_{n=0}^{\infty} \left(\sum\_{k=0}^{\left[\frac{q}{2}\right]} \binom{n}{2k}\_{p,q} (-1)^k p^{(2k-1)k} y^{2k} \,\_\circ B\_{n-k,p,q}(x,y)\right) \frac{t^n}{[n]\_{p,q}!} \tag{61}$$

$$= \frac{t}{\varepsilon\_{p,q}(t) - 1} \epsilon\_{p,q}(tx) \left(\text{CO}\_{p,q}(ty) \sin\_{p,q}(ty) - \text{SIN}\_{p,q}(ty) \cos\_{p,q}(ty)\right)$$

From Equation (61), we complete the proof of Theorem 8.

**Corollary 8.** *Putting y* = 1 *in Theorem 8, we have the following:*

$$\begin{aligned} &\sum\_{k=0}^{\left[\frac{n-1}{2}\right]} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_{p,q} (-1)^k p^{(2k+1)k} \circ B\_{n-(2k+1),p,q}(\mathbf{x}, 1) \\ &= \sum\_{k=0}^{\left[\frac{q}{2}\right]} \begin{bmatrix} n \\ 2k \end{bmatrix}\_{p,q} (-1)^k p^{(2k-1)k} \circ B\_{n-k,p,q}(\mathbf{x}, 1), \end{aligned} \tag{62}$$

*where* [*x*] *is the greatest integer not exceeding x.*

**Corollary 9.** *Setting p* = 1 *in Theorem 8, the following holds:*

$$\sum\_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \begin{bmatrix} n \\ 2k+1 \end{bmatrix}\_q (-1)^k y^{2k+1} \, \_\subset B\_{n-(2k+1),q}(\mathbf{x}, y) \quad = \sum\_{k=0}^{\lfloor \frac{n}{2} \rfloor} \begin{bmatrix} n \\ 2k \end{bmatrix}\_q (-1)^k y^{2k} \, \_\subset B\_{n-k,q}(\mathbf{x}, y), \tag{63}$$

*where <sup>C</sup>Bn*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) *is the q-cosine Bernoulli polynomials and <sup>S</sup>Bn*,*<sup>q</sup>*(*<sup>x</sup>*, *y*) *is the q-sine Bernoulli polynomials.*

**Corollary 10.** *Let p* = 1 *and q* → 1 *in Theorem 8. Then one holds*

$$\sum\_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{2k+1} (-1)^k y^{2k+1} \, \_\subset B\_{n-(2k+1)} (x, y) \quad = \sum\_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} (-1)^k y^{2k} \, \_\subset B\_{n-k} (x, y), \tag{64}$$

*where <sup>C</sup>Bn*(*<sup>x</sup>*, *y*) *is the cosine Bernoulli polynomials and <sup>S</sup>Bn*(*<sup>x</sup>*, *y*) *is the sine Bernoulli polynomials.*

### **4. Several Symmetric Properties of the (***p***,** *q***)-Cosine and (***p***,** *q***)-Sine Bernoulli Polynomials**

In this section, we point out several symmetric identities of the (*p*, *q*)-cosine and (*p*, *q*)-Bernoulli polynomials. Using various forms that are made by *a* and *b*, we obtain a few desired results regarding the (*p*, *q*)-cosine and (*p*, *q*)-sine Bernoulli polynomials. Moreover, we discover other relations of different Bernoulli polynomials by considering certain conditions in theorems. We also find the symmetric structure of the approximate roots based on the symmetric polynomials.

**Theorem 9.** *Let a and b be nonzero. Then, we obtain*

$$\begin{split} (i) \quad & \sum\_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}\_{p,q} a^{n-k-1} b^{k-1} \circ B\_{n-k,p,q} \left( \frac{\mathbf{x}}{a'}, \frac{\mathbf{y}}{a} \right) \circ B\_{k,p,q} \left( \frac{\mathbf{X}}{b'}, \frac{\mathbf{Y}}{b} \right) \\ & \quad = \sum\_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}\_{p,q} b^{n-k-1} a^{k-1} \circ B\_{n-k,p,q} \left( \frac{\mathbf{x}}{b'}, \frac{\mathbf{y}}{b} \right) \circ B\_{k,p,q} \left( \frac{\mathbf{X}}{a'}, \frac{\mathbf{Y}}{a} \right), \\ (ii) \quad & \sum\_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}\_{p,q} a^{n-k-1} b^{k-1} \circ B\_{n-k,p,q} \left( \frac{\mathbf{x}}{a'}, \frac{\mathbf{y}}{a} \right) \circ B\_{k,p,q} \left( \frac{\mathbf{X}}{b'}, \frac{\mathbf{Y}}{b} \right) \\ & \quad = \sum\_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}\_{p,q} b^{n-k-1} \circ B\_{n-k,p,q} \left( \frac{\mathbf{x}}{b'}, \frac{\mathbf{y}}{b} \right) \circ B\_{k,p,q} \left( \frac{\mathbf{X}}{a'}, \frac{\mathbf{y}}{a} \right). \end{split} \tag{65}$$
