Dtm Solutions Of The System Of Linear Differential Equations By Differential Transform Method.docx

Table of Contents Contents 2. Definition and Theorems .......................................................................................................................... 3 2.1. Two-Dimensional Differential Transform Method (DTM) ............................................................... 3 2.2. Two-Dimensional Inverse Differential Transform Method (DTM) .................................................. 3 2.3. Three-Dimensional Differential Transform Method (DTM) ............................................................. 3 2.4. Three-Dimensional Inverse Differential Transform Method (DTM) ................................................ 4 2.5. Basic Theorems of the Two-Dimensional Differential Transform Method (DTM) .......................... 4

Chapter 2 2. Definition and Theorems The basic definition 2.1. Two-Dimensional Differential Transform Method (DTM) The two-dimensional differential transform of the function m(x,y) is defined as

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

(1) 0 ,𝑦0 )

Where m(x,y) is the original function and M (a,b) is the transformed function. The M (a,b) is called respectively.

2.2. Two-Dimensional Inverse Differential Transform Method (DTM) The two-dimensional inverse differential transform of the function M(a,b) is defined as ∞

∞

𝑚(𝑥, 𝑦) = ∑ ∑ 𝑀(𝑎, 𝑏)(𝑥 − 𝑥0 )𝑎 (𝑦 − 𝑦𝑜 )𝑏

(2)

𝑎=0 𝑏=0

Equation (2) can be written as ∞

∞

𝑚(𝑥, 𝑦) = ∑ ∑ 𝑎=0 𝑏=0

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

(𝑥 − 𝑥0 )𝑎 (𝑦 − 𝑦𝑜 )𝑏

(3)

0 ,𝑦0 )

When (𝑥0, 𝑦0 ) is taken as (0,0) then equation (3) becomes ∞

∞

1 𝜕 𝑎+𝑏 𝑚(𝑥, 𝑦) = ∑ ∑ [ 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥 𝑎=0 𝑏=0

𝑥𝑎 𝑦𝑏

(4)

0 ,𝑦0 )

Where m(x,y) is the original function and M(a,b) is the transformed function and the small and capital alphabets represent the original and transformed functions respectively.

2.3. Three-Dimensional Differential Transform Method (DTM) The three-dimensional differential transform of the function f(x,y,z) is defined as

𝑀(𝑎, 𝑏, 𝑐) =

1 𝜕 𝑎+𝑏+𝑐 [ 𝑎 𝑏 𝑐 𝑚(𝑥, 𝑦, 𝑧)] 𝑎! 𝑏! 𝑐! 𝜕𝑥 𝜕𝑦 𝜕𝑡 (𝑥

(5) 0 ,𝑦0, ,𝑧0 )

Where m(x,y,z) is the original function and M(a,b,c) is the transformed function. The M(a,b,c) is called T-function and the small and capital alphabets represent the original and transformed functions respectively. 2.4. Three-Dimensional Inverse Differential Transform Method (DTM) The three-dimensional inverse differential transform of the function M(a,b,c) is defined as ∞

∞

∞

𝑚(𝑥, 𝑦) = ∑ ∑ ∑ 𝑀(𝑎, 𝑏)(𝑥 − 𝑥0 )𝑎 (𝑦 − 𝑦𝑜 )𝑏 (𝑧 − 𝑧0 )𝑐

(6)

𝑎=0 𝑏=0 𝑐=0

Equation (6) can be written as ∞

∞

∞

1 𝜕 𝑎+𝑏+𝑐 𝑚(𝑥, 𝑦) = ∑ ∑ ∑ [ 𝑚(𝑥, 𝑦, 𝑧)] 𝑎! 𝑏! 𝑐! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝜕𝑧 𝑐 (𝑥 𝑎=0 𝑏=0 𝑐=0

− 𝑦0 )𝑏 (𝑧 − 𝑧0 )𝑐

(𝑥 − 𝑥0 )𝑎 (𝑦

(7)

0 ,𝑦0 ,𝑧0 )

When (𝑥0, 𝑦0, 𝑧0 ) is taken as (0,0,0) then equation (7) becomes ∞

∞

∞

𝑚(𝑥, 𝑦) = ∑ ∑ ∑ 𝑎=0 𝑏=0 𝑐=0

1 𝜕 𝑎+𝑏+𝑐 [ 𝑎 𝑏 𝑐 𝑚(𝑥, 𝑦, 𝑧)] 𝑎! 𝑏! 𝑐! 𝜕𝑥 𝜕𝑦 𝜕𝑧 (𝑥

𝑥𝑎 𝑦𝑏 𝑧𝑐

(8)

0 ,𝑦0 ,𝑧0 )

Where m(x,y,z) is the original function and M(a,b,c) is the transformed function and the small and capital alphabets represent the original and transformed functions respectively.

2.5. Basic Theorems of the Two-Dimensional Differential Transform Method (DTM) Theorem 1 Prove that the transform function of 𝒎(𝒙, 𝒚) = 𝒍(𝒙, 𝒚) ± 𝒏(𝒙, 𝒚) Is 𝑴(𝒎, 𝒏) = 𝑳(𝒎, 𝒏) ± 𝑵(𝒎, 𝒏) Proof: By using the definition of DTM 1 𝜕 𝑎+𝑏 𝑀(𝑚, 𝑛) = [ 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ { 𝑙(𝑥, 𝑦) ± 𝑛(𝑥, 𝑦)}] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥 𝑀(𝑎, 𝑏) =

𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑙(𝑥, 𝑦) ± 𝑎 𝑏 𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 (𝑥

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

± 0 ,𝑦0 )

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

Thus we can write 𝑀(𝑎, 𝑏) = 𝐿(𝑎, 𝑏) ± 𝑁(𝑎, 𝑏)

Theorem 2 Prove that the transform function of 𝒎(𝒙, 𝒚) = 𝜷𝒍(𝒙, 𝒚) Is 𝑴(𝒂, 𝒃) = 𝜷𝑳(𝒂, 𝒃) Where 𝜷 is an arbitrary constant Proof: By using the definition of DTM 𝑀(𝑎, 𝑏) =

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝛽𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [𝛽 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

𝑀(𝑎, 𝑏) = 𝛽

0 ,𝑦0 )

Thus we can write 𝑀(𝑎, 𝑏) = 𝛽𝐿(𝑎, 𝑏)

0 ,𝑦0 )

Theorem 3 Prove that the transform function of 𝒎(𝒙, 𝒚) =

𝝏 𝒍(𝒙, 𝒚) 𝝏𝒙

Is 𝑴(𝒂, 𝒃) = (𝒂 + 𝟏)𝑳(𝒂 + 𝟏, 𝒃) Proof: By using the definition of DTM 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝑀(𝑎, 𝑏) = [ 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝜕𝑥 (𝑥 𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕 (𝑎+1)+𝑏 [ 𝑎+1 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

Multiplying and Dividing by (a+1) 𝑀(𝑎, 𝑏) =

(𝑎 + 1) 𝜕 (𝑎+1)+𝑏 [ 𝑎+1 𝑏 𝑚(𝑥, 𝑦)] (𝑎 + 1)𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 (𝑎+1)+𝑏 𝑀(𝑎, 𝑏) = (𝑎 + 1) [ 𝑚(𝑥, 𝑦)] (𝑎 + 1)! 𝑏! 𝜕𝑥 𝑎+1 𝜕𝑦 𝑏 (𝑥 Thus we can write 𝑀(𝑎, 𝑏) = (𝑎 + 1)𝐿(𝑎 + 1, 𝑏)

Theorem 4 Prove that the transform function of 𝒎(𝒙, 𝒚) =

𝝏 𝒍(𝒙, 𝒚) 𝝏𝒚

Is 𝑴(𝒂, 𝒃) = (𝒂 + 𝟏)𝑳(𝒂, 𝒃 + 𝟏) Proof:

0 ,𝑦0 )

By using the definition of DTM 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝑀(𝑎, 𝑏) = [ 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝜕𝑦 (𝑥 𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕 𝑎+(𝑏+1) [ 𝑎 𝑏+1 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

Multiplying and Dividing by (b+1) 𝑀(𝑎, 𝑏) =

(𝑏 + 1) 𝜕 𝑎+(𝑏+1) [ 𝑎 𝑏+1 𝑚(𝑥, 𝑦)] 𝑚! (𝑏 + 1)𝑛! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 (𝑎+1)+𝑏 𝑀(𝑎, 𝑏) = (𝑏 + 1) [ 𝑚(𝑥, 𝑦)] 𝑎! (𝑏 + 1)! 𝜕𝑥 𝑎+1 𝜕𝑦 𝑏 (𝑥

0 ,𝑦0 )

Thus we can write 𝑀(𝑎, 𝑏) = (𝑏 + 1)𝐿(𝑎, 𝑏 + 1)

Theorem 5 Prove that the transform function of 𝒇(𝒙, 𝒚) =

𝝏𝒓+𝒔 𝒍(𝒙, 𝒚) 𝝏𝒙𝒓 𝝏𝒚𝒔

Is 𝑴(𝒂, 𝒃) = (𝒂 + 𝟏)(𝒂 + 𝟐)(𝒂 + 𝟑) … (𝒂 + 𝒓)(𝒃 + 𝟏)(𝒃 + 𝟐)(𝒃 + 𝟑) … (𝒃 + 𝒔)𝑳(𝒂 + 𝒓, 𝒃 + 𝒔) Proof: By using the definition of DTM 1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝑟+𝑠 𝑀(𝑎, 𝑏) = [ 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝜕𝑥 𝑟 𝜕𝑦 𝑠 (𝑥

0 ,𝑦0 )

1 𝜕 (𝑎+𝑟)+(𝑏+𝑠) 𝑀(𝑎, 𝑏) = [ 𝑙(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎+𝑟 𝜕𝑦 𝑏+𝑠 (𝑥

0 ,𝑦0 )

Multiplying and Dividing by (a+r)… (a+3)(a+2)(a+1)(b+s)… (b+3)(b+2)(b+1) 𝑀(𝑎, 𝑏) (𝑎 + 𝑟) … (𝑎 + 3)(𝑎 + 2)(𝑎 + 1)(𝑏 + 𝑠) … (𝑏 + 3)(𝑏 + 2)(𝑏 + 1) 𝜕 (𝑎+𝑟)+(𝑏+𝑠) = [ 𝑙(𝑥, 𝑦)] (𝑎 + 𝑟) … (𝑎 + 3)(𝑎 + 2)(𝑎 + 1)𝑎! (𝑏 + 𝑠) … (𝑏 + 3)(𝑏 + 2)(𝑏 + 1)𝑏! 𝜕𝑥 𝑎+𝑟 𝜕𝑦 𝑏+𝑠 (𝑥 𝑀(𝑎, 𝑏) (𝑎 + 𝑟) … (𝑎 + 3)(𝑎 + 2)(𝑎 + 1)(𝑏 + 𝑠) … (𝑏 + 3)(𝑏 + 2)(𝑏 + 1) 𝜕 (𝑎+𝑟)+(𝑏+𝑠) = [ 𝑎+𝑟 𝑏+𝑠 𝑙(𝑥, 𝑦)] (𝑎 + 𝑟 )! (𝑏 + 𝑠 )! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

Thus we can write 𝑀(𝑎, 𝑏) = (𝑎 + 𝑟) … (𝑎 + 3)(𝑎 + 2)(𝑎 + 1)(𝑏 + 𝑠) … (𝑏 + 3)(𝑏 + 2)(𝑏 + 1)𝐿(𝑎 + 𝑟, 𝑏 + 𝑠) Rearranging the terms 𝑀(𝑎, 𝑏) = (𝑎 + 1)(𝑎 + 2)(𝑎 + 3) … (𝑎 + 𝑟)(𝑏 + 1)(𝑏 + 2)(𝑏 + 3) … (𝑏 + 𝑠)𝐿(𝑎 + 𝑟, 𝑏 + 𝑠)

Theorem 6 Prove that the transform function of 𝒎(𝒙, 𝒚) = 𝒍(𝒙, 𝒚)𝒏(𝒙, 𝒚) Is 𝒂

𝒃

𝑴(𝒂, 𝒃) = ∑ ∑ 𝑳(𝒓, 𝒃 − 𝒔)𝑵(𝒂 − 𝒓, 𝒔) 𝒓=𝟎 𝒔=𝟎

Proof: By using the definition of DTM 𝑀(𝑎, 𝑏) =

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕𝑎 𝜕𝑏 𝑀(𝑎, 𝑏) = [ 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥 Differentiating w.r.t ‘’y’’

0 ,𝑦0 )

0 ,𝑦0 )

1 𝜕 𝑎 𝜕 𝑏−1 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦)}] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−1 𝑦 (𝑥

0 ,𝑦0 )

Again differentiating w.r.t “y” 1 𝜕 𝑎 𝜕 𝑏−2 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)}] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−2 𝑦𝑦 (𝑥

0 ,𝑦0 )

Now differentiating w.r.t “x” 1 𝜕 𝑎−1 𝜕 𝑏−2 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎−1 𝜕𝑦 𝑏−2 𝑥𝑦𝑦 + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)

(9)

+ 𝑙(𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=1 and b=2 then 𝑀(1,2) =

1 [𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 1! 2! 𝑥𝑦𝑦 + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥 ,𝑦 ) 0

𝑀(1,2) =

0

1 1 1 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 1! 2! 𝑥𝑦 1 1 1 +2 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 1! 2!

Thus we can write 𝑀(1,2) = 𝐿(1,2)𝑁(0,0) + 𝐿(0,2)𝑁(1,0) + 𝐿(1,1)𝑁(0,1) + 𝐿(0,1)𝑁(1,1) + 𝐿(1,0)𝑁(1,1) + 𝐿(0,0)𝑁(1,2) 1

2

𝑀(1,2) = ∑ ∑ 𝐿(𝑟, 2 − 𝑝)𝑁(1 − 𝑟, 𝑝) 𝑟=0 𝑠=0

Again differentiating equation (9) w.r.t “x” 1 𝜕 𝑎−2 𝜕 𝑏−2 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−2 𝜕𝑦 𝑏−2 𝑥𝑥𝑦𝑦 + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑎−2 𝜕 𝑏−2 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 2𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−2 𝜕𝑦 𝑏−2 𝑥𝑥𝑦𝑦 + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 4𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=2 and b=2 then 𝑀(2,2) =

1 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 2𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) [𝑙 2! 2! 𝑥𝑥𝑦𝑦 + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 4𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥 ,𝑦 0

𝑀(2,2) =

0)

1 1 1 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦) + 2 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) 2! 2! 2! 2! 2! 2! 𝑦𝑦 1 1 +2 𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 4 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑦 1 1 +2 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑥 1 1 +2 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2!

Thus we can write 𝑀(2,2) = 𝐿(2,2)𝑁(0,0) + 𝐿(1,2)𝑁(1,0) + 𝐿(0,2)𝑁(2,0) + 𝐿(2,1)𝑁(0,1) + 𝐿(1,1)𝑁(1,1) + 𝐿(0,1)𝑁(2,1) + 𝐿(2,0)𝑁(0,2) + 𝐿(1,0)𝑁(1,2) + 𝐿(0,0)𝑁(2,2) 2

2

𝑀(2,2) = ∑ ∑ 𝐿(𝑟, 2 − 𝑝)𝑁(2 − 𝑟, 𝑝) 𝑟=0 𝑠=0

Similarly 3

3

𝑀(3,3) = ∑ ∑ 𝐿(𝑟, 3 − 𝑝)𝑁(3 − 𝑟, 𝑝) 𝑟=0 𝑠=0 4

4

𝑀(4,4) = ∑ ∑ 𝐿(𝑟, 4 − 𝑝)𝑁(4 − 𝑟, 𝑝) 𝑟=0 𝑠=0

And in general, we have 𝑎

𝑏

𝑀(𝑎, 𝑏) = ∑ ∑ 𝐿(𝑟, 𝑏 − 𝑠)𝑁(𝑎 − 𝑟, 𝑠) 𝑟=0 𝑠=0

Theorem 7 Prove that the transform function of 𝒎(𝒙, 𝒚) =

𝝏 𝝏 𝒍(𝒙, 𝒚) 𝒏(𝒙, 𝒚) 𝝏𝒙 𝝏𝒙

Is 𝒂

𝒃

𝒎(𝒂, 𝒃) = ∑ ∑(𝒓 + 𝟏)(𝒂 − 𝒓 + 𝟏)𝑳(𝒓 + 𝟏, 𝒃 − 𝒔)𝑵(𝒂 − 𝒓 + 𝟏, 𝒔) 𝒓=𝟎 𝒔=𝟎

Proof: By using the definition of DTM 1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥 𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝜕 [ 𝑎 𝑏 𝑙(𝑥, 𝑦) 𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 (𝑥

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

0 ,𝑦0 )

1 𝜕𝑎 𝜕𝑏 𝑀(𝑎, 𝑏) = [ 𝑙 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝑥 (𝑥

0 ,𝑦0 )

Differentiating w.r.t “y” 1 𝜕 𝑎 𝜕 𝑏−1 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)}] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−1 𝑥𝑦 (𝑥

0 ,𝑦0 )

Again differentiating w.r.t “y” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−2 [ 𝑎 𝑏−2 {𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑎 𝜕 𝑏−2 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−2 𝑥𝑦𝑦 + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Now differentiating w.r.t “x” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎−1 𝜕 𝑏−2 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−1 𝜕𝑦 𝑏−2 𝑥𝑥𝑦𝑦 + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦)

(10)

+ 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=1 and b=2 then 𝑀(1,2) =

𝑀(1,2) =

𝑀(1,2) = 2

1 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) [𝑙 1! 2! 𝑥𝑥𝑦𝑦 + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥

0 ,𝑦0 )

1 1 1 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) + 2𝑙 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 1! 2! 𝑥𝑥𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 𝑥𝑥 1 + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 1! 2! 𝑥 1 1 1 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 1! 2! 𝑥𝑥𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 1! 2! 2! 2! 𝑥𝑥 1 +2 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 2! 2! 𝑥

𝑀(1,2) = 2

1 1 1 1 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛𝑥 (𝑥, 𝑦) + 2 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 2! 2! 1! 0! 1! 2! 2! 0! 𝑥𝑥 1 1 1 1 +2 𝑙𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛𝑥𝑦 (𝑥, 𝑦) + 2 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 2! 1! 1! 1! 1! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 1 (𝑥, 𝑦) +2 𝑙𝑥𝑥 (𝑥, 𝑦) 𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2 𝑙𝑥 (𝑥, 𝑦) 𝑛 2! 0! 1! 2! 1! 0! 2! 2! 𝑥𝑥𝑦𝑦

Thus we can write 𝑀(1,2) = 2𝐿(2,2)𝑁(1,0) + 2𝐿(1,2)𝑁(2,0) + 2𝐿(2,1)𝑁(1,1) + 2𝐿(1,1)𝑁(2,1) + 2𝐿(2,0)𝑁(1,2) + 2𝐿(1,0)𝑁(2,2) 1

2

𝑀(1,2) = ∑ ∑(𝑟 + 1)(1 − 𝑟 + 1)𝐿(𝑟 + 1,2 − 𝑠)𝑁(1 − 𝑟 + 1, 𝑠) 𝑟=0 𝑠=0

Again differentiating equation (10) w .r .t “x”

1 𝜕 𝑎−2 𝜕 𝑏−2 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−2 𝜕𝑦 𝑏−2 𝑥𝑥𝑥𝑦𝑦 + 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)+2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑎−2 𝜕 𝑏−2 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−2 𝜕𝑦 𝑏−2 𝑥𝑥𝑥𝑦𝑦 + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)+4𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=2 and b=2 then 𝑀(2,2) =

1 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) [𝑙 2! 2! 𝑥𝑥𝑥𝑦𝑦 + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)+4𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥 ) 0 ,𝑦0

𝑀(2,2) =

1 1 1 𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦) + 2 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑥 (𝑥, 𝑦) 2! 2! 2! 2! 2! 2! 𝑥𝑦𝑦 1 1 + 𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) +4 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑥𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑥𝑥 1 1 +2 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥

𝑀(2,2) = 3

1 1 1 1 𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛𝑥 (𝑥, 𝑦) + 4 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 3! 2! 1! 0! 2! 2! 2! 0! 𝑥𝑥 1 1 +3 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 1! 2! 3! 0! 𝑥𝑥𝑥 1 1 1 1 +3 𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛𝑥𝑦 (𝑥, 𝑦) +4 𝑙𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 3! 1! 1! 1! 2! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 1 +3 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 3 𝑙𝑥𝑥𝑥 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 1! 1! 3! 1! 3! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 1 (𝑥, 𝑦) +4 𝑙𝑥𝑥 (𝑥, 𝑦) 𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 3 𝑙𝑥 (𝑥, 𝑦) 𝑛 2! 0! 2! 2! 1! 0! 3! 2! 𝑥𝑥𝑥𝑦𝑦

Thus we can write

𝑀(2,2) = 3𝐿(3,2)𝑁(1,0) + 4𝐿(2,2)𝑁(2,0) + 3𝐿(1,2)𝑁(3,0) + 3𝐿(3,1)𝑁(1,1) + 4𝐿(2,1)𝑁(2,1) + 3𝐿(1,1)𝑁(3,1) + 3𝐿(3,0)𝑁(1,2) + 4𝐿(2,0)𝑁(2,2) + 3𝐿(1,0)𝑁(3,2) 2

2

𝑀(2,2) = ∑ ∑(𝑟 + 1)(2 − 𝑟 + 1)𝐿(𝑟 + 1,2 − 𝑠)𝑁(2 − 𝑟 + 1, 𝑠) 𝑟=0 𝑠=0

Similarly 3

3

𝑀(3,3) = ∑ ∑(𝑟 + 1)(3 − 𝑟 + 1)𝐿(𝑟 + 1,3 − 𝑠)𝑁(3 − 𝑟 + 1, 𝑠) 𝑟=0 𝑠=0 4

4

𝑀(4,4) = ∑ ∑(𝑟 + 1)(4 − 𝑟 + 1)𝐿(𝑟 + 1,4 − 𝑠)𝑁(4 − 𝑟 + 1, 𝑠) 𝑟=0 𝑠=0

And in general, we have 𝑎

𝑏

𝑀(𝑎, 𝑏) = ∑ ∑(𝑟 + 1)(𝑎 − 𝑟 + 1)𝐿(𝑟 + 1, 𝑏 − 𝑠)𝑁(𝑎 − 𝑟 + 1, 𝑠) 𝑟=0 𝑠=0

Theorem 8 Prove that the transform function of 𝒎(𝒙, 𝒚) =

𝝏 𝝏 𝒍(𝒙, 𝒚) 𝒏(𝒙, 𝒚) 𝝏𝒚 𝝏𝒚

Is 𝒂

𝒃

𝑴(𝒂, 𝒃) = ∑ ∑(𝒔 + 𝟏)(𝒃 − 𝒔 + 𝟏)𝑳(𝒓, 𝒃 − 𝒔 + 𝟏)𝑵(𝒂 − 𝒓, 𝒔 + 𝟏) 𝒓=𝟎 𝒔=𝟎

Proof: By using the definition of DTM 1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥 𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝜕 [ 𝑎 𝑏 𝑙(𝑥, 𝑦) 𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ 𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝑦 (𝑥 𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕𝑎 𝜕𝑏 [ 𝑎 𝑏 𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

Differentiating w.r.t “y” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−1 [ 𝑎 𝑏−1 {𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)}] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

Again differentiating w.r.t “y” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−2 [ 𝑎 𝑏−2 {𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−2 [ 𝑎 𝑏−2 {𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Now differentiating w.r.t “x” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎−1 𝜕 𝑏−2 [ 𝑎−1 𝑏−2 {𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 2𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)

(11)

+ 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=1 and b=2 then 𝑀(1,2) =

𝑀(1,2) =

1 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) [𝑙 1! 2! 𝑥𝑦𝑦𝑦 + 2𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦)](𝑥

0 ,𝑦0 )

1 1 1 𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 1! 2! 𝑥𝑦𝑦 1 1 +2 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 𝑥𝑦 1 + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 1! 2! 𝑦

𝑀(1,2) = 3

1 1 1 1 𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) + 3 𝑙𝑦𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 1! 3! 0! 1! 0! 3! 1! 1! 𝑥𝑦 1 1 1 1 +4 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛𝑦𝑦 (𝑥, 𝑦) + 4 𝑙𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 1! 2! 0! 2! 0! 2! 1! 2! 𝑥𝑦𝑦 1 1 1 1 (𝑥, 𝑦) +3 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 3 𝑙𝑦 (𝑥, 𝑦) 𝑛 1! 1! 0! 3! 0! 1! 1! 3! 𝑥𝑦𝑦𝑦

Thus we can write 𝑀(1,2) = 3𝐿(1,3)𝑁(0,1) + 3𝐿(0,3)𝑁(1,1) + 4𝐿(1,2)𝑁(0,2) + 4𝐿(0,2)𝑁(1,2) + 3𝐿(1,1)𝑁(0,3) + 3𝐿(0,1)𝑁(1,3) 1

2

𝑀(𝑎, 𝑏) = ∑ ∑(𝑠 + 1)(2 − 𝑠 + 1)𝐿(𝑟, 2 − 𝑠 + 1)𝑁(1 − 𝑟, 𝑠 + 1) 𝑟=0 𝑠=0

Again differentiating equation (11) w.r.t “x” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎−2 𝜕 𝑏−2 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−2 𝜕𝑦 𝑏−2 𝑥𝑥𝑦𝑦𝑦 + 𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+2𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎−2 𝜕 𝑏−2 [ 𝑎−2 𝑏−2 {𝑙𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+4𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=2 and b=2 then 𝑀(2,2) =

1 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) [𝑙 2! 2! 𝑥𝑥𝑦𝑦𝑦 + 𝑙𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+4𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)](𝑥 ,𝑦 ) 0

0

𝑀(2,2) =

1 1 1 𝑙𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2 𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 2! 2! 𝑦𝑦𝑦 1 1 + 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) +4 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑦𝑦 1 1 +2 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑥𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑦

𝑀(2,2) = 3

1 1 1 1 𝑙𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) + 3 𝑙𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 2! 3! 0! 1! 1! 3! 1! 1! 𝑥𝑦 1 1 +3 𝑙𝑦𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 0! 3! 2! 1! 𝑥𝑥𝑦 1 1 1 1 +4 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛𝑦𝑦 (𝑥, 𝑦) +4 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 2! 2! 0! 2! 1! 2! 1! 2! 𝑥𝑦𝑦 1 1 1 1 +4 𝑙𝑦𝑦 (𝑥, 𝑦) 𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 3 𝑙𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 0! 2! 2! 2! 2! 1! 0! 3! 𝑦𝑦𝑦 1 1 1 1 (𝑥, 𝑦) +3 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 3 𝑙𝑦 (𝑥, 𝑦) 𝑛 1! 1! 1! 3! 0! 1! 2! 3! 𝑥𝑥𝑦𝑦𝑦

Thus we can write 𝑀(2,2) = 3𝐿(2,3)𝑁(0,1) + 3𝐿(1,3)𝑁(1,1) + 3𝐿(0,3)𝑁(2,1) + 4𝐿(2,2)𝑁(0,2) + 4𝐿(1,2)𝑁(1,2) + 4𝐿(0,2)𝑁(2,2) + 3𝐿(2,1)𝑁(0,3) + 3𝐿(1,1)𝑁(1,3) + 3𝐿(0,1)𝑁(2,3) 2

2

𝑀(2,2) = ∑ ∑(𝑠 + 1)(2 − 𝑠 + 1)𝐿(𝑟, 2 − 𝑠 + 1)𝑁(2 − 𝑟, 𝑠 + 1) 𝑟=0 𝑠=0

Similarly 3

3

𝑀(3,3) = ∑ ∑(𝑠 + 1)(3 − 𝑠 + 1)𝐿(𝑟, 3 − 𝑠 + 1)𝑁(3 − 𝑟, 𝑠 + 1) 𝑟=0 𝑠=0 4

4

𝑀(4,4) = ∑ ∑(𝑠 + 1)(4 − 𝑠 + 1)𝐿(𝑟, 4 − 𝑠 + 1)𝑁(4 − 𝑟, 𝑠 + 1) 𝑟=0 𝑠=0

And in general, we have 𝑎

𝑏

𝑀(𝑎, 𝑏) = ∑ ∑(𝑠 + 1)(𝑏 − 𝑠 + 1)𝐿(𝑟, 𝑏 − 𝑠 + 1)𝑁(𝑎 − 𝑟, 𝑠 + 1) 𝑟=0 𝑠=0

Theorem 9

Prove that the transform function of 𝑚(𝑥, 𝑦) =

𝜕 𝜕 𝑙(𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝜕𝑥 𝜕𝑦

Is 𝑎

𝑏

𝑀(𝑎, 𝑏) = ∑ ∑(𝑎 − 𝑟 + 1)(𝑏 − 𝑠 + 1)𝐿(𝑎 − 𝑟 + 1, 𝑠)𝑁(𝑟, 𝑏 − 𝑠 + 1) 𝑟=0 𝑠=0

Proof: By using the definition of DTM 1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥 𝑀(𝑎, 𝑏) =

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 𝜕 𝜕 [ 𝑎 𝑏 𝑙(𝑥, 𝑦) 𝑛(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 (𝑥

1 𝜕 𝑎+𝑏 𝑀(𝑎, 𝑏) = [ 𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝑥 (𝑥

0 ,𝑦0 )

0 ,𝑦0 )

1 𝜕𝑎 𝜕𝑏 𝑀(𝑎, 𝑏) = [ 𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 𝑥 (𝑥

0 ,𝑦0 )

Differentiating w.r.t “y” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−1 [ 𝑎 𝑏−1 {𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)}] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

Again differentiating w.r.t “y” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−2 [ {𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−2 𝑥𝑦𝑦 + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑎 𝜕 𝑏−2 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−2 𝑥𝑦𝑦 + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Now differentiating w.r.t “x”

1 𝜕 𝑎−1 𝜕 𝑏−2 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−1 𝜕𝑦 𝑏−2 𝑥𝑥𝑦𝑦 + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)

(12)

+ 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=1 and b=2 then 𝑀(1,2) =

𝑀(1,2) =

1 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) [𝑙 1! 2! 𝑥𝑥𝑦𝑦 + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦)](𝑥

0 ,𝑦0 )

1 1 1 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 1! 2! 𝑥𝑥𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 𝑥𝑥 1 + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 1! 2! 𝑥

𝑀(1,2) = 2

1 1 1 1 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 2! 2! 0! 1! 1! 2! 1! 1! 𝑥𝑦 1 1 1 1 +4 𝑙𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛𝑦𝑦 (𝑥, 𝑦) + 2 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 2! 1! 0! 2! 1! 1! 1! 2! 𝑥𝑦𝑦 1 1 1 1 (𝑥, 𝑦) +6 𝑙𝑥𝑥 (𝑥, 𝑦) 𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 3 𝑙𝑥 (𝑥, 𝑦) 𝑛 2! 0! 0! 3! 1! 0! 1! 3! 𝑥𝑦𝑦𝑦

Thus we can write 𝑀(1,2) = 2𝐿(2,2)𝑁(0,1) + 𝐿(1,2)𝑁(1,1) + 4𝐿(2,1)𝑁(0,2) + 2𝐿(1,1)𝑁(1,2) + 6𝐿(2,0)𝑁(0,3) + 3𝐿(1,0)𝑁(1,3) 1

2

𝑀(1,2) = ∑ ∑(1 − 𝑟 + 1)(2 − 𝑠 + 1)𝐿(1 − 𝑟 + 1, 𝑠)𝑁(𝑟, 2 − 𝑠 + 1) 𝑟=0 𝑠=0

Again differentiating equation (10) w.r.t “x” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎−2 𝜕 𝑏−2 [ 𝑎−2 𝑏−2 {𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑎−2 𝜕 𝑏−2 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑙 𝑎! 𝑏! 𝜕𝑥 𝑎−2 𝜕𝑦 𝑏−2 𝑥𝑥𝑥𝑦𝑦 + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+4𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=2 and b=2 then 𝑀(2,2) =

1 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) [𝑙 2! 2! 𝑥𝑥𝑥𝑦𝑦 + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+4𝑙𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥𝑥𝑥 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) + 2𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦)](𝑥 ,𝑦 ) 0

𝑀(2,2) =

𝑀(2,2) = 3

0

1 1 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦) 𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦) + 2 𝑙 2! 2! 2! 2! 𝑥𝑥𝑦𝑦 1 + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦 (𝑥, 𝑦) 2! 2! 𝑥𝑦𝑦 1 1 +2 𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)+4 𝑙 (𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑥𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥𝑥𝑥 1 1 +2 𝑙𝑥𝑥 (𝑥, 𝑦)𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥𝑥𝑦𝑦𝑦 (𝑥, 𝑦) 2! 2! 2! 2! 𝑥

1 1 1 1 𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) + 2 𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 3! 2! 0! 1! 2! 2! 1! 1! 𝑥𝑦 1 1 + 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 1! 2! 2! 1! 𝑥𝑥𝑦 1 1 1 1 +6 𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛𝑦𝑦 (𝑥, 𝑦)+4 𝑙𝑥𝑥𝑦 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 3! 1! 0! 2! 2! 1! 1! 2! 𝑥𝑦𝑦 1 1 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 9 𝑙𝑥𝑥𝑥 (𝑥, 𝑦) 𝑛 (𝑥, 𝑦) 1! 1! 2! 2! 3! 0! 0! 3! 𝑦𝑦𝑦 1 1 1 1 (𝑥, 𝑦) +6 𝑙𝑥𝑥 (𝑥, 𝑦) 𝑛𝑥𝑦𝑦𝑦 (𝑥, 𝑦) + 3 𝑙𝑥 (𝑥, 𝑦) 𝑛 2! 0! 1! 3! 1! 0! 2! 3! 𝑥𝑥𝑦𝑦𝑦

Thus we can write 𝑀(2,2) = 3𝐿(3,2)𝑁(0,1) + 2𝐿(2,2)𝑁(1,1) + 𝐿(1,2)𝑁(2,1) + 6𝐿(3,1)𝑁(0,2) + 4𝐿(2,1)𝑁(1,2) + 2𝐿(1,1)𝑁(2,2) + 9𝐿(3,0)𝑁(0,3) + 6𝐿(2,0)𝑁(1,3) + 3𝐿(1,0)𝑁(2,3) 2

2

𝑀(2,2) = ∑ ∑(2 − 𝑟 + 1)(2 − 𝑠 + 1)𝐿(2 − 𝑟 + 1, 𝑠)𝑁(𝑟, 2 − 𝑠 + 1) 𝑟=0 𝑠=0

Similarly 3

3

𝑀(3,3) = ∑ ∑(3 − 𝑟 + 1)(3 − 𝑠 + 1)𝐿(3 − 𝑟 + 1, 𝑠)𝑁(𝑟, 3 − 𝑠 + 1) 𝑟=0 𝑠=0 4

4

𝑀(4,4) = ∑ ∑(4 − 𝑟 + 1)(4 − 𝑠 + 1)𝐿(4 − 𝑟 + 1, 𝑠)𝑁(𝑟, 4 − 𝑠 + 1) 𝑟=0 𝑠=0

And in general, we have 𝑎

𝑏

𝑀(𝑎, 𝑏) = ∑ ∑(𝑎 − 𝑟 + 1)(𝑏 − 𝑠 + 1)𝐿(𝑎 − 𝑟 + 1, 𝑠)𝑁(𝑟, 𝑏 − 𝑠 + 1) 𝑟=0 𝑠=0

Theorem 10 Prove that the transform function of 𝒎(𝒙, 𝒚) = 𝒍(𝒙, 𝒚)𝒏(𝒙, 𝒚)𝒒(𝒙, 𝒚) Is 𝒂 𝒂−𝒓 𝒃 𝒃−𝒔

𝑴(𝒂, 𝒃) = ∑ ∑ ∑ ∑ 𝑳(𝒓, 𝒃 − 𝒔 − 𝒅)𝑵(𝒄, 𝒔)𝑸(𝒂 − 𝒓 − 𝒄, 𝒅) 𝒓=𝟎 𝒄=𝟎 𝒔=𝟎 𝒅=𝟎

Proof: By using the definition of DTM 𝑀(𝑎, 𝑏) =

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑚(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑎+𝑏 [ 𝑎 𝑏 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕𝑎 𝜕𝑏 𝑀(𝑎, 𝑏) = [ 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦)] 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏 (𝑥

0 ,𝑦0 )

Differentiating w.r.t “y” 1 𝜕 𝑎 𝜕 𝑏−1 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−1 𝑦 + 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Again differentiating w.r.t “y” 𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−2 [ 𝑎 𝑏−2 {𝑙𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝜕𝑦 + 𝑙𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 𝑙𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

𝑀(𝑎, 𝑏) =

1 𝜕 𝑎 𝜕 𝑏−2 [ {𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎 𝜕𝑦 𝑏−2 𝑦𝑦 + 2𝑙𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Now differentiating w.r.t “x” 1 𝜕 𝑎−1 𝜕 𝑏−2 𝑀(𝑎, 𝑏) = [ {𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦) 𝑎! 𝑏! 𝜕𝑥 𝑎−1 𝜕𝑦 𝑏−2 𝑥𝑦𝑦 + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2𝑙𝑥 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙(𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put a=1 and b=2 then

(13)

𝑀(1,2) =

1 [𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) 1! 2! 𝑥𝑦𝑦 + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2𝑙𝑥𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2𝑙𝑥 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙(𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑥𝑦 (𝑥, 𝑦) + 𝑙𝑥 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥 ,𝑦 ) 0

𝑀(1,2) =

0

1 1 𝑙𝑥𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞(𝑥, 𝑦) 1! 2! 1! 2! 𝑦𝑦 1 1 + 𝑙𝑦𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) 1! 2! 1! 2! 𝑥𝑦 1 1 +2 𝑙𝑦 (𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) 1! 2! 1! 2! 𝑦 1 1 +2 𝑙𝑥𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2 𝑙 (𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 𝑦 1 1 +2 𝑙𝑦 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) 1! 2! 1! 2! 𝑥 1 1 + 𝑙(𝑥, 𝑦)𝑛𝑥𝑦𝑦 (𝑥, 𝑦)𝑞(𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛𝑦𝑦 (𝑥, 𝑦)𝑞𝑥 (𝑥, 𝑦) 1! 2! 1! 2! 1 1 +2 𝑙𝑥 (𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) + 2 𝑙(𝑥, 𝑦)𝑛𝑥𝑦 (𝑥, 𝑦)𝑞𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 1 1 +2 𝑙(𝑥, 𝑦)𝑛𝑦 (𝑥, 𝑦)𝑞𝑥𝑦 (𝑥, 𝑦) + 𝑙 (𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2! 𝑥 1 1 + 𝑙(𝑥, 𝑦)𝑛𝑥 (𝑥, 𝑦)𝑞𝑦𝑦 (𝑥, 𝑦) + 𝑙(𝑥, 𝑦)𝑛(𝑥, 𝑦)𝑞𝑥𝑦𝑦 (𝑥, 𝑦) 1! 2! 1! 2!

𝑀(1,2) =

1 1 1 1 1 1 𝑙𝑥𝑦𝑦 (𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝑞(𝑥, 𝑦) + 𝑙𝑦𝑦 (𝑥, 𝑦) 𝑛𝑥 (𝑥, 𝑦) 𝑞(𝑥, 𝑦) 1! 2! 0! 0! 0! 0! 0! 2! 1! 0! 0! 0! 1 1 1 + 𝑙𝑦𝑦 (𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 2! 0! 0! 1! 0! 𝑥 1 1 1 + 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) 𝑞(𝑥, 𝑦) 1! 1! 0! 1! 0! 0! 1 1 1 + 𝑙𝑦 (𝑥, 𝑦) 𝑛𝑥𝑦 (𝑥, 𝑦) 𝑞(𝑥, 𝑦) 0! 1! 1! 1! 0! 0! 1 1 1 + 𝑙𝑦 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 1! 0! 1! 1! 0! 𝑥 1 1 1 + 𝑙𝑥𝑦 (𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 1! 1! 0! 0! 0! 1! 𝑦 1 1 1 + 𝑙𝑦 (𝑥, 𝑦) 𝑛𝑥 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 1! 1! 0! 0! 1! 𝑦 1 1 1 + 𝑙𝑦 (𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 1! 0! 0! 1! 1! 𝑥𝑦 1 1 1 + 𝑙𝑥 (𝑥, 𝑦) 𝑛𝑦𝑦 (𝑥, 𝑦) 𝑞(𝑥, 𝑦) 1! 0! 0! 2! 0! 0! 1 1 1 + 𝑙(𝑥, 𝑦) 𝑛𝑥𝑦𝑦 (𝑥, 𝑦) 𝑞(𝑥, 𝑦) 0! 0! 1! 2! 0! 0! 1 1 1 + 𝑙(𝑥, 𝑦) 𝑛𝑦𝑦 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 0! 0! 2! 1! 0! 𝑥 1 1 1 + 𝑙𝑥 (𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 1! 0! 0! 1! 0! 1! 𝑦 1 1 1 + 𝑙(𝑥, 𝑦) 𝑛𝑥𝑦 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 0! 1! 1! 0! 1! 𝑦 1 1 1 + 𝑙(𝑥, 𝑦) 𝑛𝑦 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 0! 0! 1! 1! 1! 𝑥𝑦 1 1 1 + 𝑙𝑥 (𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 1! 0! 0! 0! 0! 2! 𝑦𝑦 1 1 1 + 𝑙(𝑥, 𝑦) 𝑛𝑥 (𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 0! 1! 0! 0! 2! 𝑦𝑦 1 1 1 + 𝑙(𝑥, 𝑦) 𝑛(𝑥, 𝑦) 𝑞 (𝑥, 𝑦) 0! 0! 0! 0! 1! 2! 𝑥𝑦𝑦

Thus we can write 𝑀(1,2) = 𝐿(1,2)𝑁(0,0)𝑄(0,0) + 𝐿(0,2)𝑁(1,0)𝑄(0,0) + 𝐿(0,2)𝑁(0,0)𝑄(1,0) + 𝐿(1,1)𝑁(0,1)𝑄(0,0) + 𝐿(0,1)𝑁(1,1)𝑄(0,0) + 𝐿(0,1)𝑁(0,1)𝑄(1,0) + 𝐿(1,1)𝑁(0,0)𝑄(0,1) + 𝐿(0,1)𝑁(1,0)𝑄(0,1) + 𝐿(0,1)𝑁(0,0)𝑄(1,1) + 𝐿(1,0)𝑁(0,2)𝑄(0,0) + 𝐿(0,0)𝑁(1,2)𝑄(0,0) + 𝐿(0,0)𝑁(0,2)𝑄(1,0) + 𝐿(1,0)𝑁(0,1)𝑄(0,1) + 𝐿(0,0)𝑁(1,1)𝑄(0,1) + 𝐿(0,0)𝑁(0,1)𝑄(1,1) + 𝐿(1,0)𝑁(0,0)𝑄(0,2) + 𝐿(0,0)𝑁(1,0)𝑄(0,2) + 𝐿(0,0)𝑁(0,0)𝑄(1,2)

1 1−𝑟 2 2−𝑠

𝑀(1,2) = ∑ ∑ ∑ ∑ 𝐿(𝑟, 2 − 𝑠 − 𝑑)𝑁(𝑐, 𝑠)𝑄(1 − 𝑟 − 𝑐, 𝑑) 𝑟=0 𝑐=0 𝑠=0 𝑑=0

Again differentiating equation (13) w.r.t “x” 1 𝜕 𝑎−2 𝜕 𝑛−2 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) 𝑀(𝑎, 𝑏) = [ {𝑔 𝑎! 𝑏! 𝜕𝑥 𝑚−2 𝜕𝑦 𝑛−2 𝑥𝑥𝑦𝑦 + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑚−2 𝜕 𝑛−2 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) 𝐹(𝑚, 𝑛) = [ {𝑔 𝑚! 𝑛! 𝜕𝑥 𝑚−2 𝜕𝑦 𝑛−2 𝑥𝑥𝑦𝑦 + 2𝑥𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑥) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑥) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put m=2 and n=2 then

𝐹(2,2) =

1 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑥(𝑥, 𝑦) + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) [𝑔 2! 2! 𝑥𝑥𝑦𝑦 + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙(𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑦 (𝑥, 𝑦) + 4𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑦𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥

0 ,𝑦0 )

𝐹(2,2) =

1 1 1 1 1 1 𝑔𝑥𝑥𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙(𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 2! 2! 0! 0! 0! 0! 1! 2! 1! 0! 0! 0! 1 1 1 + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 0! 0! 1! 0! 𝑥 1 1 1 + 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 0! 2! 2! 0! 0! 0! 1 1 1 + 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 1! 0! 1! 0! 𝑥 1 1 1 + 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 0! 0! 2! 0! 𝑥𝑥 1 1 1 + 𝑔𝑥𝑥𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 2! 1! 0! 1! 0! 0! 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 1! 1! 1! 1! 0! 0! 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 0! 1! 1! 0! 𝑥 1 1 1 + 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 0! 1! 2! 1! 0! 0! 1 1 1 + 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 1! 1! 0! 𝑥 1 1 1 + 𝑔𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 0! 1! 2! 0! 𝑥𝑥 1 1 1 + 𝑔𝑥𝑥𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 1! 0! 0! 0! 1! 𝑦 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 0! 0! 1! 𝑦 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 0! 0! 1! 1! 𝑥𝑦 1 1 1 + 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 0! 0! 1! 𝑦 1 1 1 + 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 0! 1! 1! 𝑥𝑦 1 1 1 + 𝑔𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 0! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 + 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 2! 0! 0! 2! 0! 0! 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 1! 0! 1! 2! 0! 0! 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 0! 2! 1! 0! 𝑥 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙(𝑥, 𝑦) 0! 0! 2! 2! 0! 0! 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 2! 1! 0! 𝑥 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 0! 2! 0! 2! 𝑥𝑥

1 1 1 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 0! 1! 0! 1! 𝑦 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 1! 0! 1! 𝑦 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 0! 1! 1! 1! 𝑥𝑦 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 1! 0! 1! 𝑦 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 1! 1! 1! 𝑥𝑦 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 0! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 + 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 0! 0! 0! 2! 𝑦𝑦 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 0! 0! 2! 𝑦𝑦 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 0! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 0! 0! 2! 𝑦𝑦 1 1 1 + 𝑔(𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 + 𝑔(𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 0! 0! 2! 2! 𝑥𝑥𝑦𝑦 +

Thus we can write 𝐹(2,2) = 𝐺(2,2)𝐻(0,0)𝐿(0,0) + 𝐺(1,2)𝐻(1,0)𝐿(0,0) + 𝐺(1,2)𝐻(0,0)𝐿(1,0) + 𝐺(0,2)𝐻(2,0)𝐿(0,0) + 𝐺(0,2)𝐻(1,0)𝐿(1,0) + 𝐺(0,2)𝐻(0,0)𝐿(2,0) + 𝐺(2,1)𝐻(0,1)𝐿(0,0) + 𝐺(1,1)𝐻(1,1)𝐿(0,0) + 𝐺(1,1)𝐻(0,1)𝐿(1,0) + 𝐺(0,1)𝐻(2,1)𝐿(0,0) + 𝐺(0,1)𝐻(1,1)𝐿(1,0) + 𝐺(0,1)𝐻(0,1)𝐿(2,0) + 𝐺(2,1)𝐻(0,0)𝐿(0,1) + 𝐺(1,1)𝐻(1,0)𝐿(0,1) + 𝐺(1,1)𝐻(0,0)𝐿(1,1) + 𝐺(0,1)𝐻(2,0)𝐿(0,1) + 𝐺(0,1)𝐻(1,0)𝐿(1,1) + 𝐺(0,1)𝐻(0,0)𝐿(2,1) + 𝐺(2,0)𝐻(0,2)𝐿(0,0) + 𝐺(1,0)𝐻(1,2)𝐿(0,0) + 𝐺(1,0)𝐻(0,2)𝐿(1,0) + 𝐺(0,0)𝐻(2,2)𝐿(0,0) + 𝐺(0,0)𝐻(1,2)𝐿(1,0) + 𝐺(0,0)𝐻(0,2)𝐿(0,2) + 𝐺(2,0)𝐻(0,1)𝐿(0,1) + 𝐺(1,0)𝐻(1,1)𝐿(0,1) + 𝐺(1,0)𝐻(0,1)𝐿(1,1) + 𝐺(0,0)𝐻(2,1)𝐿(0,1) + 𝐺(0,0)𝐻(1,1)𝐿(1,1) + 𝐺(0,0)𝐻(0,1)𝐿(2,1) + 𝐺(2,0)𝐻(0,0)𝐿(0,2) + 𝐺(1,0)𝐻(1,0)𝐿(0,2) + 𝐺(1,0)𝐻(0,0)𝐿(1,2) + 𝐺(0,0)𝐻(2,0)𝐿(0,2) + 𝐺(0,0)𝐻(1,0)𝐿(1,2) + 𝐺(0,0)𝐻(0,0)𝐿(2,2) 2 2−𝑢 2

2−𝑣

𝐹(2,2) = ∑ ∑ ∑ ∑ 𝐺(𝑢, 2 − 𝑣 − 𝑤)𝐻(𝑡, 𝑣)𝐿(2 − 𝑢 − 𝑡, 𝑤) 𝑢=0 𝑡=0 𝑣=0 𝑤=0

Similarly

3 3−𝑢 3

3−𝑣

𝐹(3,3) = ∑ ∑ ∑ ∑ 𝐺(𝑢, 3 − 𝑣 − 𝑤)𝐻(𝑡, 𝑣)𝐿(3 − 𝑢 − 𝑡, 𝑤) 𝑢=0 𝑡=0 𝑣=0 𝑤=0 4 4−𝑢 4

4−𝑣

𝐹(4,4) = ∑ ∑ ∑ ∑ 𝐺(𝑢, 4 − 𝑣 − 𝑤)𝐻(𝑡, 𝑣)𝐿(4 − 𝑢 − 𝑡, 𝑤) 𝑢=0 𝑡=0 𝑣=0 𝑤=0

And in general, we have 𝑚 𝑚−𝑢 𝑛 𝑛−𝑣

𝐹(𝑚, 𝑛) = ∑ ∑ ∑ ∑ 𝐺(𝑢, 𝑛 − 𝑣 − 𝑤)𝐻(𝑡, 𝑣)𝐿(𝑚 − 𝑢 − 𝑡, 𝑤) 𝑢=0 𝑡=0 𝑣=0 𝑤=0

Theorem 11 Prove that the transform function of 𝒇(𝒙, 𝒚) = 𝒈(𝒙, 𝒚)

𝝏𝒉(𝒙, 𝒚) 𝝏𝒍(𝒙, 𝒚) 𝝏𝒙 𝝏𝒙

Is 𝒎 𝒎−𝒕 𝒏 𝒏−𝒗

𝑭(𝒎, 𝒏) = ∑ ∑ ∑ ∑ 𝒕=𝟎 𝒖=𝟎 𝒗=𝟎 𝒘=𝟎

(𝒖 + 𝟏)(𝒎 − 𝒕 − 𝒖 + 𝟏)𝑮(𝒕, 𝒏 − 𝒗 − 𝒘)𝑯(𝒖 + 𝟏, 𝒗) 𝑳(𝒎 − 𝒕 − 𝒖 + 𝟏, 𝒘)

Proof: By using the definition of DTM 1 𝜕 𝑚+𝑛 𝐹(𝑚, 𝑛) = [ 𝑓(𝑥, 𝑦)] 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛 (𝑥 𝐹(𝑚, 𝑛) =

𝐹(𝑚, 𝑛) =

𝐹(𝑚, 𝑛) =

Differentiating w.r.t ‘’y’’

0 ,𝑦0 )

1 𝜕 𝑚+𝑛 𝜕ℎ(𝑥, 𝑦) 𝜕𝑙(𝑥, 𝑦) [ 𝑚 𝑛 𝑔(𝑥, 𝑦) ] 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 (𝑥 1 𝜕 𝑚+𝑛 [ 𝑚 𝑛 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦)] 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

0 ,𝑦0 )

1 𝜕𝑚 𝜕𝑛 [ 𝑚 𝑛 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦)] 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑚 𝜕 𝑛−1 𝐹(𝑚, 𝑛) = [ {𝑔 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛−1 𝑦 + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Again differentiating w.r.t “y” 𝐹(𝑚, 𝑛) =

1 𝜕 𝑚 𝜕 𝑛−2 [ 𝑚 𝑛−2 {𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 + 𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

𝐹(𝑚, 𝑛) =

1 𝜕 𝑚 𝜕 𝑛−2 [ 𝑚 𝑛−2 {𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Now differentiating w.r.t “x” 𝐹(𝑚, 𝑛) =

1 𝜕 𝑚−1 𝜕 𝑛−2 [ 𝑚−1 𝑛−2 {𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑥)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put m=1 and n=2 then

(14)

𝐹(1,2) =

1 [𝑔 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 1! 2! 𝑥𝑦𝑦 + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥

0 ,𝑦0 )

𝐹(1,2) =

1 1 1 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 1! 0! 1! 0! 𝑥 1 1 1 +2 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 2! 0! 1! 0! 𝑥 1 1 1 +2 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑥) 0! 2! 1! 0! 2! 0! 𝑥𝑥 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 1! 1! 0! 𝑥 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 1! 1! 0! 𝑥 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 1! 2! 0! 𝑥𝑥 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 0! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 0! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 2! 1! 0! 𝑥 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 2! 1! 0! 𝑥 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 2! 2! 0! 𝑥𝑥 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 1! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑥) 0! 0! 2! 1! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 + 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔(𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 0! 0! 1! 0! 2! 2! 𝑥𝑥𝑦𝑦

Thus we can write 𝐹(1,2) = 𝐺(1,2)𝐻(1,0)𝐿(1,0) + 2𝐺(0,2)𝐻(2,0)𝐿(1,0) + 2𝐺(0,2)𝐻(1,0)𝐿(2,0) + 𝐺(1,1)𝐻(1,1)𝐿(1,0) + 2𝐺(0,1)𝐻(2,1)𝐿(1,0) + 2𝐺(0,1)𝐻(1,1)𝐿(2,0) + 𝐺(1,1)𝐻(1,0)𝐿(1,1) + 2𝐺(0,1)𝐻(2,0)𝐿(1,1) + 2𝐺(0,1)𝐻(1,0)𝐿(2,1) + 𝐺(1,0)𝐻(1,2)𝐿(1,0) + 2𝐺(0,0)𝐻(2,2)𝐿(1,0) + 2𝐺(0,0)𝐻(1,2)𝐿(2,0) + 𝐺(1,0)𝐻(1,1)𝐿(1,1) + 2𝐺(0,0)𝐻(2,1)𝐿(1,1) + 2𝐺(0,0)𝐻(1,1)𝐿(2,1) + 𝐺(1,0)𝐻(1,0)𝐿(1,2) + 2𝐺(0,0)𝐻(2,0)𝐿(1,2) + 2𝐺(0,0)𝐻(1,0)𝐿(2,2)

1 1−𝑡 2

2−𝑣

𝐹(1,2) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(𝑢 + 1)(1 − 𝑥 − 𝑢 + 1)𝐺(𝑡, 2 − 𝑣 − 𝑤)𝐻(𝑢 + 1, 𝑣) 𝐿(1 − 𝑡 − 𝑢 + 1, 𝑤)

Again differentiating equation (14) w.r.t “x” 1 𝜕 𝑚−2 𝜕 𝑛−2 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 𝐹(𝑚, 𝑛) = [ {𝑔 𝑚! 𝑛! 𝜕𝑥 𝑚−2 𝜕𝑦 𝑛−2 𝑥𝑥𝑦𝑦 + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑚−2 𝜕 𝑛−2 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) 𝐹(𝑚, 𝑛) = [ {𝑔 𝑚! 𝑛! 𝜕𝑥 𝑥−2 𝜕𝑦 𝑛−2 𝑥𝑥𝑦𝑦 + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝒙𝟎 ,𝒚𝟎 )

Put m=2 and n=2 then

𝐹(2,2) =

1 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) [𝑔 2! 2! 𝑥𝑥𝑦𝑦 + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑥𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑦 (𝑥, 𝑦) + 4𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑥)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝒙

𝟎 ,𝒚𝟎 )

𝐹(2,2) =

1 1 1 𝑔𝑥𝑥𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 2! 1! 0! 1! 0! 𝑥 1 1 1 +2 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 2! 0! 1! 0! 𝑥 1 1 1 +2 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 1! 0! 2! 0! 𝑥𝑥 1 1 1 +3 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 3! 0! 1! 0! 𝑥 1 1 1 +4 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 2! 0! 2! 0! 𝑥𝑥 1 1 1 +3 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 1! 0! 3! 0! 𝑥𝑥𝑥 1 1 1 + 𝑔𝑥𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 1! 1! 1! 1! 0! 𝑥 1 1 1 +2 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 2! 1! 1! 0! 𝑥 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 1! 2! 0! 𝑥𝑥 1 1 1 +3 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 3! 1! 1! 0! 𝑥 1 1 1 +4 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 1! 2! 0! 𝑥𝑥 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 1! 2! 0! 𝑥𝑥 1 1 1 +3 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 1! 3! 0! 𝑥𝑥𝑥 1 1 1 + 𝑔𝑥𝑥𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 1! 1! 0! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 2! 0! 1! 1! 𝑥𝑦 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 +3 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 3! 0! 1! 1! 𝑥𝑦 1 1 1 +4 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 + 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +3 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 0! 1! 1! 0! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 + 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 1! 2! 1! 0! 𝑥 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 2! 2! 1! 0! 𝑥 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 2! 2! 0! 𝑥𝑥

1 1 1 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 3! 2! 1! 0! 𝑥 1 1 1 +4 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 2! 2! 0! 𝑥𝑥 1 1 1 +3 𝑔(𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 2! 3! 0! 𝑥𝑥𝑥 1 1 1 + 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 1! 1! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 2! 1! 1! 1! 𝑥𝑦 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 +3 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 3! 1! 1! 1! 𝑥𝑦 1 1 1 +4 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +3 𝑔(𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 0! 0! 1! 1! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 + 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 1! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 2! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 1! 0! 1! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 +3 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 3! 0! 1! 2! 𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +4 𝑔(𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 0! 0! 2! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +3 𝑔(𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 0! 0! 1! 0! 3! 2! 𝑥𝑥𝑥𝑦𝑦 +3

Thus we can write

𝐹(2,2) = 𝐺(2,2)𝐻(1,0)𝐿(1,0) + 2𝐺(1,2)𝐻(2,0)𝐿(1,0) + 2𝐺(1,2)𝐻(1,0)𝐿(2,0) + 3𝐺(0,2)𝐻(3,0)𝐿(1,0) + 4𝐺(0,2)𝐻(2,0)𝐿(2,0) + 3𝐺(0,2)𝐻(1,0)𝐿(3,0) + 𝐺(2,1)𝐻(1,1)𝐿(1,0) + 2𝐺(1,1)𝐻(2,1)𝐿(1,0) + 𝐺(1,1)𝐻(1,1)𝐿(2,0) + 3𝐺(0,1)𝐻(3,1)𝐿(1,0) + 4𝐺(0,1)𝐻(2,1)𝐿(2,0) + 𝐺(1,1)𝐻(1,1)𝐿(2,0) + 3𝐺(0,1)𝐻(1,1)𝐿(3,0) + 𝐺(2,1)𝐻(1,0)𝐿(1,1) + 2𝐺(1,1)𝐻(2,0)𝐿(1,1) + 𝐺(1,1)𝐻(1,0)𝐿(2,1) + 3𝐺(0,1)𝐻(3,0)𝐿(1,1) + 4𝐺(0,1)𝐻(2,0)𝐿(2,1) + 𝐺(1,1)𝐻(1,0)𝐿(2,1) + 3𝐺(0,1)𝐻(1,0)𝐿(3,1) + 𝐺(2,0)𝐻(1,2)𝐿(1,0) + 2𝐺(1,0)𝐻(2,2)𝐿(1,0) + 2𝐺(1,0)𝐻(1,2)𝐿(2,0) + 3𝐺(0,0)𝐻(3,2)𝐿(1,0) + 4𝐺(0,0)𝐻(2,2)𝐿(2,0) + 3𝐺(0,0)𝐻(1,2)𝐿(3,0) + 𝐺(2,0)𝐻(1,1)𝐿(1,1) + 2𝐺(1,0)𝐻(2,1)𝐿(1,1) + 2𝐺(1,0)𝐻(1,1)𝐿(2,1) + 3𝐺(0,0)𝐻(3,1)𝐿(1,1) + 4𝐺(0,0)𝐻(2,1)𝐿(2,1) + 3𝐺(0,0)𝐻(1,1)𝐿(3,1) + 𝐺(2,0)𝐻(1,0)𝐿(1,2) + 2𝐺(1,0)𝐻(2,0)𝐿(1,2) + 2𝐺(1,0)𝐻(1,0)𝐿(2,2) + 3𝐺(0,0)𝐻(3,0)𝐿(1,2) + 4𝐺(0,0)𝐻(2,0)𝐿(2,2) + 3𝐺(0,0)𝐻(1,0)𝐿(3,2) 2 2−𝑥 2

2−𝑣

𝐹(2,2) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(𝑢 + 1)(2 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 2 − 𝑣 − 𝑤)𝐻(𝑢 + 1, 𝑣) 𝐿(2 − 𝑡 − 𝑢 + 1, 𝑤)

Similarly 3 3−𝑡 3

3−𝑣

𝐹(3,3) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0 4 4−𝑡 4

4−𝑣

𝐹(4,4) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(𝑢 + 1)(3 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 3 − 𝑣 − 𝑤)𝐻(𝑢 + 1, 𝑣) 𝐿(3 − 𝑡 − 𝑢 + 1, 𝑤) (𝑢 + 1)(4 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 4 − 𝑣 − 𝑤)𝐻(𝑢 + 1, 𝑣) 𝐿(4 − 𝑡 − 𝑢 + 1, 𝑤)

And in general, we have 𝑚 𝑚−𝑡 𝑛 𝑛−𝑣

𝐹(𝑚, 𝑛) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(𝑢 + 1)(𝑚 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 𝑛 − 𝑣 − 𝑤)𝐻(𝑢 + 1, 𝑣) 𝐿(𝑚 − 𝑡 − 𝑢 + 1, 𝑤)

Theorem 12 Prove that the transform function of 𝒇(𝒙, 𝒚) = 𝒈(𝒙, 𝒚)𝒉(𝒙, 𝒚)

𝝏𝟐 𝒍(𝒙, 𝒚) 𝝏𝒙𝟐

Is 𝒎 𝒎−𝒕 𝒏 𝒏−𝒗

𝑭(𝒎, 𝒏) = ∑ ∑ ∑ ∑ 𝒕=𝟎 𝒖=𝟎 𝒗=𝟎 𝒘=𝟎

Proof:

(𝒎 − 𝒕 − 𝒖 + 𝟐)(𝒎 − 𝒕 − 𝒖 + 𝟏)𝑮(𝒕, 𝒏 − 𝒗 − 𝒘)𝑯(𝒖, 𝒗) 𝑳(𝒎 − 𝒕 − 𝒖 + 𝟐, 𝒘)

By using the definition of Differential Transform Method 𝐹(𝑚, 𝑛) =

1 𝜕 𝑚+𝑛 [ 𝑚 𝑛 𝑓(𝑥, 𝑦)] 𝑚! 𝑛! 𝜕𝑥 𝜕𝑦 (𝑥

0 ,𝑦0 )

1 𝜕 𝑚+𝑛 𝜕 2 𝑙(𝑥, 𝑦) 𝐹(𝑚, 𝑛) = [ 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦) ] 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛 𝜕𝑥 2 (𝑥 1 𝜕 𝑚+𝑛 𝐹(𝑚, 𝑛) = [ 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦)] 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛 (𝑥

0 ,𝑦0 )

0 ,𝑦0 )

1 𝜕𝑚 𝜕𝑛 𝐹(𝑚, 𝑛) = [ 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦)] 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛 (𝑥

0 ,𝑦0 )

Differentiating w.r.t ‘’y’’ 1 𝜕 𝑚 𝜕 𝑛−1 𝐹(𝑚, 𝑛) = [ {𝑔 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛−1 𝑦 + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Again differentiating w.r.t “y” 1 𝜕 𝑚 𝜕 𝑛−2 𝐹(𝑚, 𝑛) = [ {𝑔 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛−2 𝑦𝑦 + 𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑥)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑚 𝜕 𝑛−2 𝐹(𝑚, 𝑛) = [ {𝑔 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝑚 𝜕𝑦 𝑛−2 𝑦𝑦 + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Now differentiating w.r.t “x”

1 𝜕 𝑚−1 𝜕 𝑛−2 𝐹(𝑚, 𝑛) = [ {𝑔 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 𝑚! 𝑛! 𝜕𝑥 𝑚−1 𝜕𝑦 𝑛−2 𝑥𝑦𝑦 + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑥)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦)

15

+ 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

Put m=1 and n=2 then 𝐹(1,2) =

1 [𝑔 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 1! 2! 𝑥𝑦𝑦 + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝑥 ,𝑦 0

0)

𝐹(1,2) = 2

Thus we can write

1 1 1 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 0! 0! 2! 0! 𝑥𝑥 1 1 1 +2 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 1! 0! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 0! 0! 3! 0! 𝑥𝑥𝑥 1 1 1 +2 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 0! 1! 2! 0! 𝑥𝑥 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 1! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 0! 1! 3! 0! 𝑥𝑥𝑥 1 1 1 +2 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 0! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 0! 1! 0! 0! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 2! 0! 2! 0! 𝑥𝑥 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 2! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔(𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 0! 3! 0! 𝑥𝑥𝑥 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 0! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔(𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 0! 0! 0! 1! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔𝑥 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 1! 0! 0! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔(𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 0! 0! 1! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔(𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 0! 0! 0! 0! 3! 2! 𝑥𝑥𝑥𝑦𝑦

𝐹(1,2) = 2𝐺(1,2)𝐻(0,0)𝐿(2,0) + 2𝐺(0,2)𝐻(1,0)𝐿(2,0) + 6𝐺(0,2)𝐻(0,0)𝐿(3,0) + 2𝐺(1,1)𝐻(0,1)𝐿(2,0) + 2𝐺(0,1)𝐻(1,1)𝐿(2,0) + 6𝐺(0,1)𝐻(0,1)𝐿(3,0) + 2𝐺(1,1)𝐻(0,0)𝐿(2,1) + 2𝐺(0,1)𝐻(1,0)𝐿(2,1) + 6𝐺(0,1)𝐻(0,0)𝐿(3,1) + 2𝐺(1,0)𝐻(2,0)𝐿(2,0) + 2𝐺(0,0)𝐻(1,2)𝐿(2,0) + 6𝐺(0,0)𝐻(2,0)𝐿(3,0) + 2𝐺(1,0)𝐻(0,1)𝐿(2,1) + 2𝐺(0,0)𝐻(1,1)𝐿(2,1) + 6𝐺(0,0)𝐻(0,1)𝐿(3,1) + 2𝐺(1,0)𝐻(0,0)𝐿(2,2) + 2𝐺(0,0)𝐻(1,0)𝐿(2,2) + 6𝐺(0,0)𝐻(0,0)𝐿(3,2) 1 1−𝑡

2

2−𝑣

𝐹(1,2) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(1 − 𝑡 − 𝑢 + 2)(1 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 2 − 𝑣 − 𝑤)𝐻(𝑢, 𝑣) 𝐿(1 − 𝑡 − 𝑢 + 2, 𝑤)

Now differentiating equation (15) w.r.t “x”

1 𝜕 𝑚−2 𝜕 𝑛−2 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 𝐹(𝑚, 𝑛) = [ {𝑔 𝑚! 𝑛! 𝜕𝑥 𝑚−2 𝜕𝑦 𝑛−2 𝑥𝑥𝑦𝑦 + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑥)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝑥0 ,𝑦0 )

1 𝜕 𝑚−2 𝜕 𝑛−2 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) 𝐹(𝑚, 𝑛) = [ {𝑔 𝑚! 𝑛! 𝜕𝑥 𝑚−2 𝜕𝑦 𝑛−2 𝑥𝑥𝑦𝑦 + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)}] (𝒙𝟎 ,𝒚𝟎 )

Put m=2 and n=2 then

𝐹(2,2) =

1 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) [𝑔 2! 2! 𝑥𝑥𝑦𝑦 + 2𝑔𝑥𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔𝑦𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑦 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔𝑦 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦𝑥 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑦𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥 (𝑥, 𝑦) + 2𝑔𝑥𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔𝑥 (𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑦 (𝑥, 𝑦) + 4𝑔(𝑥, 𝑦)ℎ𝑥𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑦 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦 (𝑥, 𝑦) + 𝑔𝑥𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔𝑥 (𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ𝑥𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 2𝑔(𝑥, 𝑦)ℎ𝑥 (𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦) + 𝑔(𝑥, 𝑦)ℎ(𝑥, 𝑦)𝑙𝑥𝑥𝑥𝑥𝑦𝑦 (𝑥, 𝑦)](𝒙

𝟎 ,𝒚𝟎 )

𝐹(2,2) = 2

1 1 1 𝑔𝑥𝑥𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 2! 0! 0! 2! 0! 𝑥𝑥 1 1 1 +2 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 1! 0! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔𝑥𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 2! 0! 0! 3! 0! 𝑥𝑥𝑥 1 1 1 +2 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 2! 0! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 2! 1! 0! 3! 0! 𝑥𝑥𝑥 1 1 1 (𝑥, 𝑦) + 12 𝑔𝑦𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 0! 2! 0! 0! 4! 0! 𝑥𝑥𝑥𝑥 1 1 1 +2 𝑔𝑥𝑥𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 1! 0! 1! 2! 0! 𝑥𝑥 1 1 1 +2 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 1! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 0! 1! 3! 0! 𝑥𝑥𝑥 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 1! 2! 0! 𝑥𝑥 1 1 1 +6 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 1! 1! 3! 0! 𝑥𝑥𝑥 1 1 1 (𝑥, 𝑦) + 12 𝑔𝑦 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 0! 1! 0! 1! 4! 0! 𝑥𝑥𝑥𝑥 1 1 1 +2 𝑔𝑥𝑥𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 1! 0! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 +2 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 1! 1! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔𝑥𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 1! 1! 0! 0! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 +2 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 1! 2! 0! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔𝑦 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 0! 1! 1! 0! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) + 12 𝑔𝑦 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 0! 1! 0! 0! 4! 1! 𝑥𝑥𝑥𝑥𝑦 1 1 1 +2 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 0! 2! 2! 0! 𝑥𝑥 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 2! 2! 0! 𝑥𝑥 1 1 1 +6 𝑥 (𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 𝑥 0! 2! 3! 0! 𝑥𝑥𝑥 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 2! 2! 0! 𝑥𝑥

1 1 1 𝑔(𝑥, 𝑦) ℎ𝑥𝑦𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 1! 2! 3! 0! 𝑥𝑥𝑥 1 1 1 (𝑥, 𝑦) + 12 𝑔(𝑥, 𝑦) ℎ𝑦𝑦 (𝑥, 𝑦) 𝑙 0! 0! 0! 2! 4! 0! 𝑥𝑥𝑥𝑥 1 1 1 +2 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 2! 0! 0! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 1! 0! 1! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔𝑥 (𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 1! 0! 0! 1! 3! 1! 𝑥𝑥𝑥𝑦 1 1 1 +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑥𝑦 (𝑥, 𝑦) 𝑙 (𝑥, 𝑦) 0! 0! 2! 1! 2! 1! 𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔(𝑥, 𝑦) ℎ𝑥𝑦 (𝑥, 𝑦) 𝑙 0! 0! 1! 1! 3! 0! 𝑥𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) + 12 𝑔(𝑥, 𝑦) ℎ𝑦 (𝑥, 𝑦) 𝑙 0! 0! 0! 1! 4! 1! 𝑥𝑥𝑥𝑥𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔𝑥𝑥 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 2! 0! 0! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔𝑥 (𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 1! 0! 1! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔𝑥 (𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 1! 0! 0! 0! 3! 2! 𝑥𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +2 𝑔(𝑥, 𝑦) ℎ𝑥𝑥 (𝑥, 𝑦) 𝑙 0! 0! 2! 0! 2! 2! 𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) +6 𝑔(𝑥, 𝑦) ℎ𝑥 (𝑥, 𝑦) 𝑙 0! 0! 1! 0! 3! 2! 𝑥𝑥𝑥𝑦𝑦 1 1 1 (𝑥, 𝑦) + 12 𝑔(𝑥, 𝑦) ℎ(𝑥, 𝑦) 𝑙 0! 0! 0! 0! 4! 2! 𝑥𝑥𝑥𝑥𝑦𝑦 +6

Thus we can write 𝐹(2,2) = 2𝐺(2,2)𝐻(0,0)𝐿(2,0) + 2𝐺(1,2)𝐻(1,0)𝐿(2,0) + 6𝐺(1,2)𝐻(0,0)𝐿(3,0) + 2𝐺(0,2)𝐻(2,0)𝐿(2,0) + 6𝐺(0,2)𝐻(1,0)𝐿(3,0) + 12𝐺(0,2)𝐻(0,0)𝐿(4,0) + 2𝐺(2,1)𝐻(0,1)𝐿(2,0) + 2𝐺(1,1)𝐻(1,1)𝐿(2,0) + 6𝐺(1,1)𝐻(0,1)𝐿(3,0) + 2𝐺(0,1)𝐻(2,1)𝐿(2,0) + 6𝐺(0,1)𝐻(1,1)𝐿(3,0) + 12𝐺(0,1)𝐻(0,1)𝐿(4,0) + 2𝐺(2,1)𝐻(0,0)𝐿(2,1) + 2𝐺(1,1)𝐻(1,0)𝐿(2,1) + 6𝐺(1,1)𝐻(0,0)𝐿(3,1) + 2𝐺(0,1)𝐻(2,0)𝐿(2,1) + 6𝐺(0,1)𝐻(1,0)𝑥(3,1) + 12𝐺(0,1)𝐻(0,0)𝐿(4,1) + 2𝐺(2,0)𝐻(0,2)𝐿(2,0) + 2𝐺(1,0)𝐻(1,2)𝐿(2,0) + 6𝐺(1,0)𝐻(0,2)𝐿(3,0) + 2𝐺(0,0)𝐻(2,2)𝐿(2,0) + 6𝐺(0,0)𝐻(1,2)𝐿(3,0) + 12𝐺(0,0)𝐻(0,2)𝐿(4,0) + 2𝐺(2,0)𝐻(0,1)𝐿(2,1) + 2𝐺(1,0)𝐻(1,1)𝐿(2,1) + 6𝐺(1,0)𝐻(0,1)𝐿(3,1) + 2𝐺(0,0)𝐻(2,1)𝐿(2,1) + 6𝐺(0,0)𝐻(1,1)𝐿(3,0) + 12𝐺(0,0)𝐻(0,1)𝐿(4,1) + 2𝐺(2,0)𝐻(0,0)𝐿(2,2) + 2𝐺(1,0)𝐻(1,0)𝐿(2,2) + 6𝐺(1,0)𝐻(0,0)𝐿(3,2) + 2𝐺(0,0)𝐻(2,0)𝐿(2,2) + 6𝐺(0,0)𝐻(1,0)𝐿(3,2) + 12𝐺(0,0)𝐻(0,0)𝐿(4,2)

2 2−𝑡

2

2−𝑣

𝐹(2,2) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(2 − 𝑡 − 𝑢 + 2)(2 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 2 − 𝑣 − 𝑤)𝐻(𝑢, 𝑣) 𝐿(2 − 𝑡 − 𝑢 + 2, 𝑤)

Similarly 3 3−𝑡

3

3−𝑣

𝐹(3,3) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

4 4−𝑡

4

4−𝑣

𝐹(4,4) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(3 − 𝑡 − 𝑢 + 2)(3 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 3 − 𝑣 − 𝑤)𝐻(𝑢, 𝑣) 𝐿(3 − 𝑡 − 𝑢 + 2, 𝑤)

(4 − 𝑡 − 𝑢 + 2)(4 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 4 − 𝑣 − 𝑤)𝐻(𝑢, 𝑣) 𝐿(4 − 𝑡 − 𝑢 + 2, 𝑤)

And in general, we have 𝑚 𝑚−𝑡 𝑛 𝑛−𝑣

𝐹(𝑚, 𝑛) = ∑ ∑ ∑ ∑ 𝑡=0 𝑢=0 𝑣=0 𝑤=0

(𝑚 − 𝑡 − 𝑢 + 2)(𝑚 − 𝑡 − 𝑢 + 1)𝐺(𝑡, 𝑛 − 𝑣 − 𝑤)𝐻(𝑢, 𝑣) 𝐿(𝑚 − 𝑡 − 𝑢 + 2, 𝑤)

Theorem 13 If 𝒇(𝒙, 𝒚) = 𝒙𝒖 𝒚𝒗 Then 𝑭(𝒎, 𝒏) = 𝜹(𝒎 − 𝒖, 𝒏 − 𝒗)