3. Special polynomials

To obtain the polynomials described in the previous section, one can use different methods, some tougher than others. These polynomials are typically obtained as a result of the solution of each specific differential equation by means of the power series method. Usually, it is also shown that they can be obtained through a generating function and also by using the Rodrigues formula for each special polynomial, or finally, through a contour integral. Most Mathematical Methods courses also include a study of the properties of these polynomials, such as orthogonality, completeness, recursion relations, special values, asymptotic expansions and their relation to other functions, such as polynomials and hypergeometric functions. There is no doubt that this is a challenging and demanding subject that requires a great deal of attention from most students.

#### 3.1 Differential equation

The most common way to solve the special polynomials is solving the associated differential equation through power series and the Frobenius method <sup>y</sup> <sup>¼</sup> <sup>∑</sup><sup>∞</sup> <sup>n</sup>¼<sup>0</sup>anx<sup>n</sup>. The corresponding polynomials satisfy the following differential equations:

the Laguerre differential equation,

$$
\infty \mathbf{y''} + (\mathbf{1} - \mathbf{x}) \mathbf{y'} + n \mathbf{y} = \mathbf{0},
\tag{28}
$$

the Hermite differential equation,

$$
\mathbf{y}' - 2\mathbf{x}\mathbf{y}' + 2\mathbf{n}\mathbf{y} = \mathbf{0},\tag{29}
$$

the Legendre differential equation,

$$(\mathbf{1} - \mathbf{x}^2)\mathbf{y}'' - 2\mathbf{x}\mathbf{y}' + n(n+\mathbf{1})\mathbf{y} = \mathbf{0} \,, \tag{30}$$

the Tchebycheff differential equation,

$$(\mathbf{1} - \mathbf{x}^2)\mathbf{y}'' - \mathbf{x}\mathbf{y}' + n^2\mathbf{y} = \mathbf{0},\tag{31}$$

and the Gegenbauer differential equation,

$$(\mathbf{1} - \mathbf{x}^2)y'' - (2\lambda + \mathbf{1})\mathbf{x}y' + n(n + 2\lambda)y = \mathbf{0},\tag{32}$$

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

with <sup>n</sup> <sup>¼</sup> <sup>0</sup>, <sup>1</sup>, <sup>2</sup>, <sup>3</sup>, … in all the previous cases. Note that if <sup>λ</sup> <sup>¼</sup> <sup>1</sup> , Eq. (32) reduces <sup>2</sup> to the Legendre differential equation (Eq. (30)), and if λ ¼ 0, Eq. (32) reduces to the Tchebycheff differential equation (Eq. (31)).

#### 3.2 Rodrigues formula

For polynomials ψnð Þ x , with interval I, weight function w xð Þ, and an eigenvalue equation of the form

$$p(\mathbf{x})\boldsymbol{\upmu}\_{n}^{\prime\prime}(\mathbf{x}) + q(\boldsymbol{\upx})\boldsymbol{\upmu}\_{n}^{\prime}(\mathbf{x}) + \boldsymbol{\uplambda}\_{n}\boldsymbol{\upmu}\_{n}(\mathbf{x}) = \mathbf{0},\tag{33}$$

<sup>0</sup> <sup>ð</sup> ð Þw xð ÞÞ and with q xð Þ¼ p x , the general formula w xð Þ

$$\boldsymbol{w}\_{n}(\mathbf{x}) = \boldsymbol{w}(\boldsymbol{\mathfrak{x}})^{-1} \frac{d^{n}}{d\boldsymbol{\mathfrak{x}}^{n}} [\boldsymbol{p}(\boldsymbol{\mathfrak{x}})^{n}\boldsymbol{w}(\boldsymbol{\mathfrak{x}})] \tag{34}$$

is known as the Rodrigues formula, useful to obtain the nth-degree polynomial of ψ [15].

#### 3.3 Generating function and contour integral

Let Γ be a curve that encloses x∈I but excludes the endpoints of I. Then, considering the Cauchy integral formula [16] for derivatives of w xð Þp xð Þ<sup>n</sup> to derive an integral formula from Eq. (34), one obtains

$$\frac{\Psi\_n(\mathbf{x})}{n!} = \frac{1}{2\pi i} \int\_{\Gamma} \frac{w(z)}{w(\mathbf{x})} \frac{p(z)^n}{\left(z - \mathbf{x}\right)^n} \frac{dz}{z - \mathbf{x}}.\tag{35}$$

<sup>n</sup> <sup>o</sup> <sup>ψ</sup>nð Þ <sup>x</sup> The generating function for the orthogonal polynomials is defined as <sup>n</sup>!

$$G(\mathbf{x}, \mathbf{s}) = \sum\_{n=0}^{\infty} \frac{\mu\_n(\mathbf{x})}{n!} \mathbf{s}^n. \tag{36}$$

In the following section, Laguerre [2], Hermite [17], Legendre, Tchebycheff [18] and Gegenbauer [3] polynomials are obtained through a simple method, using basic linear algebra concepts, such as the eigenvalue and the eigenvector of a matrix.

### 4. Simple approach to special polynomials

The general algebraic polynomial of degree n,

$$a\_0 + a\_1\varkappa + a\_2\varkappa^2 + a\_3\varkappa^3 + \dots a\_n\varkappa^n,\tag{37}$$

with ao, a1,…,an ∈ ℜ, is represented by vector

$$A\_n = \begin{bmatrix} a\_0 \\ a\_1 \\ a\_2 \\ a\_3 \\ \vdots \\ a\_n \end{bmatrix}. \tag{38}$$

Taking the first derivative of the above polynomial (x), one obtains the polynomial

$$\frac{d}{d\mathbf{x}}\left[a\_0 + a\_1\mathbf{x} + a\_2\mathbf{x}^2 + a\_3\mathbf{x}^3 + \dots \\ a\_n\mathbf{x}^n\right] = a\_1 + 2a\_2\mathbf{x} + 3a\_3\mathbf{x}^2 + \dots \\ na\_n\mathbf{x}^{n-1},\tag{39}$$

which may be written as

$$\frac{dA\_n}{dx} = \begin{bmatrix} a\_1 \\ 2a\_2 \\ 3a\_3 \\ \vdots \\ na\_n \\ 0 \end{bmatrix}. \tag{40}$$

Taking the second derivative of the polynomial (Eq. (37)) one obtains

$$\frac{d^2}{d\mathbf{x}^2} \left[ a\_0 + a\_1 \mathbf{x} + a\_2 \mathbf{x}^2 + a\_3 \mathbf{x}^3 + \dots \mathbf{a}\_n \mathbf{x}^n \right] = 2a\_2 + 6a\_3 \mathbf{x} + \dots n(n-1)a\_n \mathbf{x}^{n-2}, \tag{41}$$

which may be written as

$$\frac{d^2A\_n}{dx^2} = \begin{bmatrix} 2a\_2 \\ 6a\_3 \\ \vdots \\ n(n-1)a\_n \\ 0 \\ 0 \end{bmatrix}. \tag{42}$$

Using Eq. (40), Eq. (39) may be written as

$$
\begin{bmatrix}
\mathbf{0} & \mathbf{1} & \mathbf{0} & \mathbf{0} & \cdots & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{2} & \mathbf{0} & \cdots & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{3} & \cdots & \mathbf{0} \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \cdots & n \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \cdots & \mathbf{0} \\
\end{bmatrix}
\begin{bmatrix}
a\_{0} \\
a\_{1} \\
a\_{2} \\
a\_{3} \\
\vdots \\
a\_{n} \\
a\_{n} \\
\end{bmatrix} = \begin{bmatrix}
a\_{1} \\
2a\_{2} \\
3a\_{3} \\
\vdots \\
na\_{n} \\
0 \\
\end{bmatrix};\tag{43}
$$

therefore, the first derivative operator An may be written as

$$\frac{d}{dx} \rightarrow \begin{bmatrix} \mathbf{0} & \mathbf{1} & \mathbf{0} & \mathbf{0} & \cdots & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & 2 & \mathbf{0} & \cdots & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{3} & \cdots & \mathbf{0} \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \cdots & n \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \cdots & \mathbf{0} \end{bmatrix}.\tag{44}$$

Doing the same for Eq. (41),

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

$$
\begin{bmatrix}
0 & 0 & 2 & 0 & \cdots & 0 \\
0 & 0 & 0 & 6 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & n(n-1) \\
0 & 0 & 0 & 0 & \cdots & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}
\begin{bmatrix}
a\_0 \\
a\_1 \\
a\_2 \\
\vdots \\
a\_{n-1} \\
a\_n \\
a\_n \\
\end{bmatrix} = \begin{bmatrix}
a\_1 \\
2a\_2 \\
\vdots \\
n(n-1)a\_n \\
0 \\
0 \\
\end{bmatrix},
\tag{45}
$$

the second derivative operator An may be written as

$$\frac{d^2}{dx^2} \rightarrow \begin{bmatrix} 0 & 0 & 2 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 6 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & n(n-1) \\ 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}.\tag{46}$$

#### 4.1 Laguerre

The Laguerre differential operator is given by.

$$
\propto \frac{d^2}{d\mathbf{x}^2} + (\mathbf{1} - \mathbf{x})\frac{d}{d\mathbf{x}};\tag{47}
$$

substituting Eqs. (41) and (44) into Eq. (47),

$$\begin{aligned} \propto \left[ 2a\_2 + 6a\_3 \mathbf{x} + \dots \right. &+ n(n-1)a\_n \mathbf{x}^{n-2} \right] + (1-\mathbf{x}) \left[ a\_1 + 2a\_2 \mathbf{x} + 3a\_3 \mathbf{x}^2 + \dots \right. &+ n a\_n \mathbf{x}^{n-1} \right] \\ &= a\_1 + (4a\_2 - a\_1)\mathbf{x} + (9a\_3 - 2a\_2)\mathbf{x}^2 + (16a\_4 + 3a\_3)\mathbf{x}^3 + \dots - n a\_n \mathbf{x}^n \end{aligned} \tag{48}$$

which may be written as

$$
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & \cdots & 0 \\
0 & -1 & 4 & 0 & 0 & \cdots & 0 \\
0 & 0 & -2 & 9 & 0 & \cdots & 0 \\
0 & 0 & 0 & -3 & 16 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & 0 & \cdots & -n \\
\end{bmatrix}
\begin{bmatrix}
a\_0 \\
a\_1 \\
a\_2 \\
a\_3 \\
\vdots \\
a\_n \\
\end{bmatrix} = \begin{bmatrix}
a\_1 \\
4a\_2 - a\_1 \\
9a\_3 - 2a\_2 \\
16a\_4 - 3a\_3 \\
\vdots \\
\end{bmatrix}.
\tag{49}
$$

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

$$\times \frac{d^2}{d\mathbf{x}^2} + (\mathbf{1} - \mathbf{x}) \frac{d}{d\mathbf{x}} \to \begin{bmatrix} \mathbf{0} & \mathbf{1} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & -\mathbf{1} & \mathbf{4} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & -\mathbf{2} & \mathbf{9} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{3} \end{bmatrix}.\tag{50}$$

The eigenvalues of a matrix M are the values that satisfy the equation DetðM � λIÞ ¼ 0. However, since the matrix (Eq. (50)) 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 �2, λ<sup>4</sup> ¼ �3. 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� :

$$
\begin{bmatrix}
\mathbf{0} - \boldsymbol{\lambda}\_i & \mathbf{1} & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & -\mathbf{1} - \boldsymbol{\lambda}\_i & \mathbf{4} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & -\mathbf{2} - \boldsymbol{\lambda}\_i & \mathbf{9} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{3} - \boldsymbol{\lambda}\_i
\end{bmatrix}
\begin{bmatrix}
a\_0 \\ a\_1 \\ a\_2 \\ a\_3
\end{bmatrix} = \begin{bmatrix}
\mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}.\tag{51}
$$

Substituting eigenvalue λ<sup>1</sup> ¼ 0 in Eq. (51), we obtain eigenvector v1:

$$\begin{aligned} \boldsymbol{v}\_1 &= \begin{bmatrix} \mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix}; \tag{52} \end{aligned} \tag{52}$$

the elements of this eigenvector correspond to the first Laguerre polynomial, L0ðxÞ ¼ 1:

Substituting eigenvalue λ<sup>2</sup> ¼ �1 in Eq. (51), we obtain eigenvector v2:

$$v\_2 = \begin{bmatrix} \mathbf{1} \\ -\mathbf{1} \\ \mathbf{0} \\ \mathbf{0} \end{bmatrix};\tag{53}$$

the elements of this eigenvector correspond to the second Laguerre polynomial, L1ðxÞ ¼ 1 � x:

Substituting eigenvalue λ<sup>3</sup> ¼ �2 in Eq. (51), we obtain eigenvector v3:

$$v\_3 = \begin{bmatrix} \mathbf{1} \\ -2 \\ \mathbf{1} \\ \mathbf{2} \\ \mathbf{0} \end{bmatrix};\tag{54}$$

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

Substituting eigenvalue λ<sup>4</sup> ¼ �3 in Eq. (51), we obtain eigenvector v4:

$$v\_4 = \begin{bmatrix} 1 \\ -3 \\ 3 \\ \frac{3}{2} \\ -\frac{1}{6} \end{bmatrix};\tag{55}$$

the elements of this eigenvector correspond to the fourth Laguerre polynomial, <sup>2</sup> � <sup>1</sup> <sup>3</sup> <sup>L</sup><sup>3</sup> <sup>x</sup> <sup>2</sup> <sup>x</sup> <sup>6</sup> <sup>ð</sup> Þ ¼ <sup>1</sup> � <sup>3</sup><sup>x</sup> <sup>þ</sup> <sup>x</sup> : <sup>3</sup>
