4.2 Hermite

The Hermite differential operator is given by

Simple Approach to Special Polynomials: Laguerre, Hermite, Legendre,Tchebycheff… DOI: http://dx.doi.org/10.5772/intechopen.83029

$$\frac{d^2}{d\mathfrak{x}^2} - 2\mathfrak{x}\frac{d}{d\mathfrak{x}};\tag{56}$$

substituting Eqs. (41) and (44) into Eq. (56),

$$\begin{aligned} \left[2a\_2 + 6a\_3\mathbf{x} + \dots + n(n-1)a\_n\mathbf{x}^{n-2}\right] - 2\mathbf{x}\left[a\_1 + 2a\_2\mathbf{x} + 3a\_3\mathbf{x}^2 + \dots + na\_n\mathbf{x}^{n-1}\right] \\ = 2a\_2 + (6a\_3 - 2a\_1)\mathbf{x} + (12a\_4 - 4a\_2)\mathbf{x}^2 + (20a\_5 - 6a\_3)\mathbf{x}^3 + \dots - 2na\_n \end{aligned} \tag{57}$$

which may be written as

$$
\begin{bmatrix}
0 & 0 & 2 & 0 & 0 & \cdots & 0 \\
0 & -2 & 0 & 6 & 0 & \cdots & 0 \\
0 & 0 & -4 & 0 & 12 & \cdots & 0 \\
0 & 0 & 0 & -6 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & 0 & \cdots & -2n
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3 \\ \vdots \\ a\_n
\end{bmatrix} = \begin{bmatrix}
2a\_2 \\ 6a\_3 - 2a\_1 \\ 12a\_4 - 4a\_2 \\ 20a\_5 - 6a\_3 \\ \vdots \\ -2na\_n
\end{bmatrix}.
\tag{58}
$$

For simplicity, the Hermite differential operator, as a 4x4 matrix, is represented by

$$\frac{d^2}{dx^2} - 2\varkappa \frac{d}{dx} \to \begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 & -2 & 0 & 6 \\ 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & -6 \end{bmatrix}.\tag{59}$$

The eigenvalues of a matrix M are the values that satisfy the equation DetðM � λIÞ ¼ 0. However, since the matrix (Eq. (59)) is a triangular matrix, the eigenvalues λ<sup>i</sup> of this matrix are the elements of the diagonal, namely: λ<sup>1</sup> ¼ 0, λ<sup>2</sup> ¼ �2, λ<sup>3</sup> ¼ �4, λ<sup>4</sup> ¼ �6. The corresponding eigenvectors are the solutions of <sup>T</sup> the equation <sup>ð</sup><sup>M</sup> � <sup>λ</sup>iIÞ � <sup>v</sup> <sup>¼</sup> 0, where the eigenvector <sup>v</sup> ¼ ½a0; <sup>a</sup>1; <sup>a</sup>2; <sup>a</sup>3� :

$$
\begin{bmatrix}
\mathbf{0} - \lambda\_i & \mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{0} & -\mathbf{2} - \lambda\_i & \mathbf{0} & \mathbf{6} \\
\mathbf{0} & \mathbf{0} & -\mathbf{4} - \lambda\_i & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{6} - \lambda\_i
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3
\end{bmatrix} = \begin{bmatrix}
\mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}.\tag{60}
$$

Substituting eigenvalue λ<sup>1</sup> ¼ 0 in Eq. (60), we obtain eigenvector v1:

$$v\_1 = \begin{bmatrix} \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{61}$$

the elements of this eigenvector correspond to the first Hermite polynomial, H0ðxÞ ¼ 1:

Substituting eigenvalue λ<sup>2</sup> ¼ �2 in Eq. (60), we obtain eigenvector v2:

$$v\_2 = \begin{bmatrix} \mathbf{0} \\ \mathbf{2} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{62}$$

the elements of this eigenvector correspond to the second Hermite polynomial, H1ðxÞ ¼ 2x:

Substituting eigenvalue λ<sup>3</sup> ¼ �4 in Eq. (60), we obtain eigenvector v3:

$$v\_3 = \begin{bmatrix} -2 \\ 0 \\ 4 \\ 0 \end{bmatrix};\tag{63}$$

the elements of this eigenvector correspond to the third Hermite polynomial, <sup>H</sup><sup>2</sup> <sup>x</sup><sup>ð</sup> Þ ¼ <sup>4</sup>x<sup>2</sup> � <sup>2</sup>:

Substituting eigenvalue λ<sup>4</sup> ¼ �6 in Eq. (60), we obtain eigenvector v4:

$$v\_4 = \begin{bmatrix} \mathbf{0} \\ -\mathbf{1}2 \\ \mathbf{0} \\ \mathbf{8} \end{bmatrix};\tag{64}$$

the elements of this eigenvector correspond to the fourth Hermite polynomial, <sup>H</sup><sup>3</sup> <sup>x</sup><sup>ð</sup> Þ ¼ <sup>8</sup>x<sup>3</sup> � <sup>12</sup>x:

#### 4.3 Legendre

The Legendre differential operator is given by

$$(1 - x^2)\frac{d^2}{d\mathbf{x}^2} - 2x\frac{d}{d\mathbf{x}};\tag{65}$$

substituting Eqs. (41) and (44) into Eq. (65),

$$\begin{aligned} \left(1 - x^2\right) \left[2a\_2 + 6a\_3x + \dots + n(n-1)a\_nx^{n-2}\right] - 2x\left[a\_1 + 2a\_2x + 3a\_3x^2 + \dots + na\_nx^{n-1}\right] \\ = 2a\_2 + (6a\_3 - 2a\_1)x + (12a\_4 - 6a\_2)x^2 + (20a\_5 - 12a\_3)x^3 + \dots - (n^2 + n)a\_n \end{aligned} \tag{66}$$

which may be written as

$$
\begin{bmatrix}
0 & 0 & 2 & 0 & 0 & \cdots & 0 \\
0 & -2 & 0 & 6 & 0 & \cdots & 0 \\
0 & 0 & -6 & 0 & 12 & \cdots & 0 \\
0 & 0 & 0 & -12 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & 0 & \cdots & -(n^2 + n)
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3 \\ \vdots \\ a\_n
\end{bmatrix} = \begin{bmatrix}
2a\_2 \\ 6a\_3 - 2a\_1 \\ 12a\_4 - 6a\_2 \\ 20a\_5 - 12a\_3 \\ \vdots \\ -(n^2 + n)a\_n
\end{bmatrix}.\tag{67}
$$

For simplicity, the Legendre differential operator, as a 4x4 matrix, is represented by

$$(1-x^2)\frac{d^2}{dx^2} - 2x\frac{d}{dx} \to \begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 & -2 & 0 & 6 \\ 0 & 0 & -6 & 0 \\ 0 & 0 & 0 & -12 \end{bmatrix}.\tag{68}$$

Simple Approach to Special Polynomials: Laguerre, Hermite, Legendre,Tchebycheff… DOI: http://dx.doi.org/10.5772/intechopen.83029

The eigenvalues of a matrix M are the values that satisfy the equation DetðM � λIÞ ¼ 0. However, since the matrix (Eq. (68)) is a triangular matrix, the eigenvalues λ<sup>i</sup> of this matrix are the elements of the diagonal, namely: λ<sup>1</sup> ¼ 0, λ<sup>2</sup> ¼ �2, λ<sup>3</sup> ¼ �6, λ<sup>4</sup> ¼ �12. The corresponding eigenvectors are the solutions of the <sup>T</sup> equation <sup>ð</sup><sup>M</sup> � <sup>λ</sup>iIÞ � <sup>v</sup> <sup>¼</sup> 0, where the eigenvector <sup>v</sup> ¼ ½a0; <sup>a</sup>1; <sup>a</sup>2; <sup>a</sup>3� :

$$
\begin{bmatrix}
\mathbf{0} - \lambda\_i & \mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{0} & -\mathbf{2} - \lambda\_i & \mathbf{0} & \mathbf{6} \\
\mathbf{0} & \mathbf{0} & -\mathbf{6} - \lambda\_i & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{1}\mathbf{2} - \lambda\_i
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3
\end{bmatrix} = \begin{bmatrix}
\mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}.\tag{69}
$$

Substituting eigenvalue λ<sup>1</sup> ¼ 0 in Eq. (69), we obtain eigenvector v1:

$$\begin{aligned} \boldsymbol{v}\_1 &= \begin{bmatrix} \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}; \tag{70} \end{aligned} \tag{70}$$

the elements of this eigenvector correspond to the first Legendre polynomial, P0ðxÞ ¼ 1:

Substituting eigenvalue λ<sup>2</sup> ¼ �2 in Eq. (69), we obtain eigenvector v2:

$$v\_2 = \begin{bmatrix} \mathbf{0} \\ \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{71}$$

the elements of this eigenvector correspond to the second Legendre polynomial, P1ðxÞ ¼ x:

Substituting eigenvalue λ<sup>3</sup> ¼ �6 in Eq. (69), we obtain eigenvector v3:

$$\boldsymbol{v}\_{3} = \begin{bmatrix} \mathbf{1} \\ \mathbf{0} \\ -\mathbf{3} \\ \mathbf{0} \end{bmatrix};\tag{72}$$

the elements of this eigenvector correspond to the third Legendre polynomial, <sup>P</sup><sup>2</sup> <sup>x</sup> <sup>2</sup> <sup>x</sup><sup>2</sup> � <sup>2</sup> <sup>1</sup> <sup>ð</sup> Þ ¼ : <sup>3</sup>

Substituting eigenvalue λ<sup>4</sup> ¼ �12 in Eq. (69), we obtain eigenvector v4:

$$v\_4 = \begin{bmatrix} \mathbf{0} \\ \mathbf{3} \\ \mathbf{0} \\ \mathbf{0} \\ -\mathbf{5} \end{bmatrix};\tag{73}$$

the elements of this eigenvector correspond to the fourth Legendre polynomial, <sup>P</sup><sup>3</sup> <sup>x</sup> <sup>2</sup> <sup>x</sup><sup>3</sup> � <sup>2</sup> <sup>3</sup> <sup>ð</sup> Þ ¼ <sup>x</sup>: <sup>5</sup>

#### 4.4 Tchebycheff

The Tchebycheff differential operator is given by

$$\left(1-\varkappa^{2}\right)\frac{d^{2}}{d\varkappa^{2}}-\varkappa\frac{d}{d\varkappa};\tag{74}$$

substituting Eqs. (41) and (44) into Eq. (74),

$$\begin{aligned} \left(1-\mathbf{x}^{2}\right)\left[2a\_{2}+\mathsf{6}a\_{3}\mathbf{x}+\ldots+n(n-1)a\_{n}\mathbf{x}^{n-2}\right] &-\mathbf{x}\left[a\_{1}+2a\_{2}\mathbf{x}+\mathbf{3}a\_{3}\mathbf{x}^{2} \\ +\ldots &+na\_{n}\mathbf{x}^{n-1}\right] &= 2a\_{2}+(\mathsf{6}a\_{3}-a\_{1})\mathbf{x}+(12a\_{4}-4a\_{2})\mathbf{x}^{2} \\ +(20a\_{5}-9a\_{3})\mathbf{x}^{3}+\cdots-n^{2}a\_{n} \end{aligned} \tag{75}$$

which may be written as

$$
\begin{bmatrix}
0 & 0 & 2 & 0 & 0 & \cdots & 0 \\
0 & -1 & 0 & 6 & 0 & \cdots & 0 \\
0 & 0 & -4 & 0 & 12 & \cdots & 0 \\
0 & 0 & 0 & -9 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & 0 & \cdots & -n^2
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3 \\ \vdots \\ a\_n
\end{bmatrix} = \begin{bmatrix}
2a\_2 \\ 6a\_3 - a\_1 \\ 12a\_4 - 4a\_2 \\ 20a\_5 - 9a\_3 \\ \vdots \\ -n^2a\_n
\end{bmatrix}.
\tag{76}
$$

For simplicity, the Tchebycheff differential operator, as a 4x4 matrix, is represented by

$$(1-x^2)\frac{d^2}{dx^2} - x\frac{d}{dx} \to \begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 & -1 & 0 & 6 \\ 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & -9 \end{bmatrix}.\tag{77}$$

The eigenvalues of a matrix M are the values that satisfy the equation DetðM � λIÞ ¼ 0. However, since the matrix (Eq. (77)) is a triangular matrix, the eigenvalues λ<sup>i</sup> of this matrix are the elements of the diagonal, namely: λ<sup>1</sup> ¼ 0, λ<sup>2</sup> ¼ �1, λ<sup>3</sup> ¼ �4, λ<sup>4</sup> ¼ �9. The corresponding eigenvectors are the solutions of the <sup>T</sup> equation <sup>ð</sup><sup>M</sup> � <sup>λ</sup>iIÞ � <sup>v</sup> <sup>¼</sup> 0, where the eigenvector <sup>v</sup> ¼ ½a0; <sup>a</sup>1; <sup>a</sup>2; <sup>a</sup>3� ;

$$
\begin{bmatrix}
\mathbf{0} - \lambda\_i & \mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{0} & -\mathbf{1} - \lambda\_i & \mathbf{0} & \mathbf{6} \\
\mathbf{0} & \mathbf{0} & -\mathbf{4} - \lambda\_i & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{9} - \lambda\_i
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3
\end{bmatrix} = 
\begin{bmatrix}
\mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0}
\end{bmatrix}.\tag{78}
$$

Substituting eigenvalue λ<sup>1</sup> ¼ 0 in Eq. (78), we obtain eigenvector v1:

$$\begin{aligned} v\_1 &= \begin{bmatrix} \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{79} \end{aligned} \tag{79}$$

the elements of this eigenvector correspond to the first Tchebycheff polynomial, T0ðxÞ ¼ 1:

Substituting eigenvalue λ<sup>2</sup> ¼ �1 in Eq. (78), we obtain eigenvector v2:

Simple Approach to Special Polynomials: Laguerre, Hermite, Legendre,Tchebycheff… DOI: http://dx.doi.org/10.5772/intechopen.83029

$$v\_2 = \begin{bmatrix} \mathbf{0} \\ \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{80}$$

the elements of this eigenvector correspond to the second Tchebycheff polynomial, T1ðxÞ ¼ x:

Substituting eigenvalue λ<sup>3</sup> ¼ �4 in Eq. (78), we obtain eigenvector v3:

$$v\_3 = \begin{bmatrix} -1 \\ 0 \\ 2 \\ 0 \end{bmatrix};\tag{81}$$

the elements of this eigenvector correspond to the third Tchebycheff polynomial, <sup>T</sup><sup>2</sup> <sup>x</sup><sup>ð</sup> Þ ¼ <sup>2</sup>x<sup>2</sup> � <sup>1</sup>:

Substituting eigenvalue λ<sup>4</sup> ¼ �9 in Eq. (78), we obtain eigenvector v4:

$$v\_4 = \begin{bmatrix} \mathbf{0} \\ -\mathbf{3} \\ \mathbf{0} \\ \mathbf{4} \end{bmatrix}. \tag{82}$$

the elements of this eigenvector correspond to the fourth Tchebycheff polynomial, <sup>T</sup><sup>3</sup> <sup>x</sup><sup>ð</sup> Þ ¼ <sup>4</sup>x<sup>3</sup> � <sup>3</sup>x:

#### 4.5 Gegenbauer

The Gegenbauer differential operator is given by

$$(\mathbf{1} - \mathbf{x}^2) \frac{d^2}{d\mathbf{x}^2} - (2\boldsymbol{\lambda} + \mathbf{1})\boldsymbol{\kappa} \frac{d}{d\mathbf{x}};\tag{83}$$

substituting (41) and (44) into (83),

$$\begin{aligned} \left(1 - \mathbf{x}^2\right) \left[2a\_2 + 6a\_3 \mathbf{x} + \dots \ + n(n-1)a\_n \mathbf{x}^{n-2}\right] - (2\lambda + 1)\mathbf{x}[a\_1 \\ + 2a\_2 \mathbf{x} + 3a\_3 \mathbf{x}^2 + \dots + na\_n \mathbf{x}^{n-1}] &= 2a\_2 + [6a\_3 - (2\lambda + 1)a\_1] \mathbf{x} \\ + \left[12a\_4 - 4(\lambda + 1)a\_2\right] \mathbf{x}^2 + [2\mathbf{0}a\_5 - 3(2\lambda + 3)a\_3] \mathbf{x}^3 \\ + \dots - \left[n^2 + 2\lambda n\right] a\_n \end{aligned} \tag{84}$$

which may be written as

$$
\begin{bmatrix}
0 & 0 & 2 & 0 & 0 & \cdots & 0 \\
0 & -(2\lambda + 1) & 0 & 6 & 0 & \cdots & 0 \\
0 & 0 & -4(\lambda + 1) & 0 & 12 & \cdots & 0 \\
0 & 0 & 0 & -3(2\lambda + 3) & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & 0 & \cdots & -n^2 - 2\lambda n
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3 \\ \vdots \\ a\_n \end{bmatrix} = \begin{bmatrix}
2a\_2 \\ 6a\_3 - (2\lambda + 1)a\_1 \\ 12a\_4 - 4(\lambda + 1)a\_2 \\ 20a\_3 - 3(2\lambda + 3)a\_3 \\ \vdots \\ -(n^2 + 2\lambda n)a\_n
\end{bmatrix}.\tag{85}
$$

Applied Mathematics

For simplicity, the Gegenbauer differential operator, as a 4x4 matrix, is represented by

$$(1 - x^2)\frac{d^2}{dx^2} - (2\lambda + 1)x\frac{d}{dx} \to \begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 & -(2\lambda + 1) & 0 & 6 \\ 0 & 0 & -4(\lambda + 1) & 0 \\ 0 & 0 & 0 & -3(2\lambda + 3) \end{bmatrix} . \tag{86}$$

� � The eigenvalues of a matrix M are the values that satisfy the equation DetðM � λ<sup>0</sup> IÞ ¼ 0. However, since the matrix (Eq. (86)) is a triangular matrix, the eigenvalues λ<sup>i</sup> of this matrix are the elements of the diagonal, namely: λ<sup>0</sup> ¼ 0, λ0 ¼ �ð2λ þ 1Þ, λ<sup>0</sup> , λ<sup>0</sup> are the solutions of the equation Þ. The corresponding eigenvectors ¼ �4ðλ þ 1Þ <sup>4</sup> ¼ �3 2ð λ þ 3 M � λ<sup>0</sup> i I � v ¼ 0, where the eigenvector v ¼ ½a0; a1; a2; a3� T ;

$$
\begin{bmatrix}
\mathbf{0} - \lambda\_i' & \mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{0} & -(2\lambda + 1) - \lambda\_i' & \mathbf{0} & \mathbf{6} \\
\mathbf{0} & \mathbf{0} & -\mathbf{4}(\lambda + 1) - \lambda\_i' & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{3}(2\lambda + 3) - \lambda\_i'
\end{bmatrix} \cdot \begin{bmatrix} a\_0 \\ a\_1 \\ a\_2 \\ a\_3 \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}.\tag{87}
$$

Substituting eigenvalue λ<sup>0</sup> ¼ 0 in Eq. (87), we obtain eigenvector v1:

$$v\_1 = \begin{bmatrix} \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{88}$$

the elements of this eigenvector correspond to the first Gegenbauer polynomial, Cλ ðxÞ ¼ 1:

Substituting eigenvalue λ<sup>0</sup> ¼ �ð2λ þ 1Þ in Eq. (87), we obtain eigenvector v2:

$$v\_2 = \begin{bmatrix} \mathbf{0} \\ \mathbf{2}\boldsymbol{\lambda} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{89}$$

the elements of this eigenvector correspond to the second Gegenbauer polynomial, C<sup>λ</sup> ðxÞ ¼ 2λx.

Substituting eigenvalue λ<sup>0</sup> ¼ �4ðλ þ 1Þ in Eq. (87), we obtain eigenvector v3:

$$v\_3 = \begin{bmatrix} -\lambda \\ 0 \\ 2\lambda(1+\lambda) \\ 0 \end{bmatrix};\tag{90}$$

the elements of this eigenvector correspond to the third Gegenbauer polynomial, Cλ <sup>x</sup> <sup>ð</sup> <sup>x</sup><sup>2</sup> <sup>ð</sup> Þ ¼ �<sup>λ</sup> <sup>þ</sup> <sup>2</sup><sup>λ</sup> <sup>1</sup> <sup>þ</sup> <sup>λ</sup><sup>Þ</sup> .

Substituting eigenvalue λ<sup>0</sup> ¼ �3 2ð λ þ 3Þ in Eq. (87), we obtain eigenvector v4: Simple Approach to Special Polynomials: Laguerre, Hermite, Legendre,Tchebycheff… DOI: http://dx.doi.org/10.5772/intechopen.83029

$$v\_4 = \begin{bmatrix} 0 \\ -2\lambda(1+\lambda) \\ 0 \\ \frac{4}{3}\lambda(1+\lambda)(2+\lambda) \end{bmatrix};\tag{91}$$

the elements of this eigenvector correspond to the fourth Gegenbauer polynomial, <sup>3</sup> <sup>C</sup><sup>λ</sup> <sup>x</sup> ð Þ<sup>x</sup> <sup>þ</sup> <sup>4</sup> <sup>ð</sup> Þ ¼ <sup>2</sup><sup>λ</sup> <sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>λ</sup>ð<sup>1</sup> <sup>þ</sup> <sup>λ</sup>Þð Þ <sup>2</sup> <sup>þ</sup> <sup>λ</sup> <sup>x</sup> . <sup>3</sup> <sup>3</sup>
