# Solution of ODE

### Contents

**Determine Ordinary and Singular Points of a Differential Equation**1**Determine Regular and Irregular Singular Points of a Differential Equation**2**Recall Power Series**4**Find the Solution of a Differential Equation using Power Series Method**6**Recall Frobenius Method**9**Find the Solution of a Differential Equation using Frobenius series method about Regular Singular Point**9**Describe Bessel Equation**9**Find the Series Solution of Bessel Equation**9**Describe Bessel Function**9**Find the Recurrence Relation of Bessel Functions**9**Recall the Orthogonal Property of Bessel Functions**9**Describe the Generating Function of Bessel Function**9**Recall Bessel’s Integral**9**Recall Legendre Equation**9**Find the series solution of Legendre Equation**9**Recall Legendre Polynomial**9**Recall Rodrigues’ Formula for Legendre Polynomial**9**Describe Generating Function of Legendre Polynomial**9**Recall Orthogonality of Legendre Polynomial**9**Recall the Recurrence Relation**9**Find the Recurrence Relation of Legendre Polynomial**9

Determine Ordinary and Singular Points of a Differential Equation

When studying differential equations, it is important to distinguish between ordinary and singular points. An ordinary point of a differential equation is a point at which the coefficients and functions in the differential equation are finite and well-behaved. In contrast, a singular point is a point at which one or more of the coefficients or functions becomes infinite or undefined.

The type of singularity at a singular point can be classified as either a regular singularity or an irregular singularity. A regular singularity is a singular point at which the coefficients and functions in the differential equation are well-behaved and can be expressed as power series in terms of the distance from the singular point. In contrast, an irregular singularity is a singular point at which the coefficients and functions in the differential equation cannot be expressed as power series in terms of the distance from the singular point.

To determine whether a point is an ordinary point or a singular point, we can examine the coefficients and functions in the differential equation. If all the coefficients and functions are finite and well-behaved at a point, then that point is an ordinary point. If one or more of the coefficients or functions becomes infinite or undefined at a point, then that point is a singular point.

Examples:

- Consider the differential equation y” + 3xy’ + 2y = 0. To determine whether the point x = 0 is an ordinary point or a singular point, we can examine the coefficients and functions in the differential equation. At x = 0, the coefficients and functions are all finite and well-behaved, so x = 0 is an ordinary point.
- Consider the differential equation x
^{2}y” + xy’ + (x^{2}– 1/4)y = 0. To determine whether the point x = 0 is an ordinary point or a singular point, we can examine the coefficients and functions in the differential equation. At x = 0, the coefficient x^{2}becomes zero, which makes the point x = 0 a singular point. To determine whether the singularity is regular or irregular, we can rewrite the differential equation as x^{2}y” + xy’ + (x^{2}– 1/4)y = 0. Factoring out x^{2}, we get x^{2}(y” + (1/x)y’ + ((x^{2}– 1/4)/x^{2})y) = 0. The coefficient of y” + (1/x)y’ + ((x^{2}– 1/4)/x^{2})y is well-behaved and can be expressed as a power series in terms of the distance from x = 0. Therefore, the singularity at x = 0 is a regular singularity.

In conclusion, distinguishing between ordinary and singular points is an important step in the study of differential equations. An ordinary point is a point at which the coefficients and functions in the differential equation are finite and well-behaved, while a singular point is a point at which one or more of the coefficients or functions becomes infinite or undefined. Understanding the type of singularity at a singular point is also important, as it can affect the methods used to solve the differential equation.

Determine Regular and Irregular Singular Points of a Differential Equation

When studying differential equations, it is important to distinguish between regular and irregular singular points. A singular point of a differential equation is a point at which one or more of the coefficients or functions becomes infinite or undefined. If the coefficients and functions in the differential equation can be expressed as power series in terms of the distance from the singular point, then the singular point is said to be a regular singular point. If the coefficients and functions cannot be expressed as power series in terms of the distance from the singular point, then the singular point is said to be an irregular singular point.

To determine whether a singular point is regular or irregular, we need to examine the behavior of the coefficients and functions in the differential equation as we approach the singular point. If the coefficients and functions can be expressed as power series in terms of the distance from the singular point, then the singular point is regular. If the coefficients and functions cannot be expressed as power series in terms of the distance from the singular point, then the singular point is irregular.

Regular Singular Points:

Regular singular points are characterized by their ability to be expressed as power series in terms of the distance from the singular point. If we can find a series solution of a differential equation around a regular singular point, then we can find all the other solutions around the same point. Regular singular points arise frequently in physical applications, such as in the solution of the Bessel’s equation and Legendre’s equation.

For example, consider the differential equation:

x^{2}y” + xy’ + (x^{2} – 1)y = 0

The point x = 0 is a regular singular point of the differential equation. To see this, we can substitute y = x^{r} into the differential equation and obtain:

r(r-1) + r – (r^{2} – 1) = 0

Simplifying the expression, we get:

r^{2} – 1 = 0

(r – 1)(r + 1) = 0

Therefore, the roots of the indicial equation are r = 1 and r = -1. Because the roots differ by an integer, x = 0 is a regular singular point.

Irregular Singular Points:

Irregular singular points are characterized by their inability to be expressed as power series in terms of the distance from the singular point. Finding solutions around irregular singular points is more complicated and requires different techniques. Irregular singular points arise less frequently in physical applications, but they are important in the study of special functions, such as the confluent hypergeometric functions and the Kummer’s equation.

For example, consider the differential equation:

x^{2}y” + xy’ + (x^{2} – x)y = 0

The point x = 0 is an irregular singular point of the differential equation. To see this, we can substitute y = x^{r} into the differential equation and obtain:

r(r-1) + r – (r^{2} – r) = 0

Simplifying the expression, we get:

r^{2} – 1 = 0

(r – 1)(r + 1) = 0

Therefore, the roots of the indicial equation are r = 1 and r = -1. Because the roots differ by an integer, x = 0 is a singular point. However, the coefficients and functions in the differential equation cannot be expressed as power series in terms of the distance from the singular point, so x = 0 is an irregular singular point.

In conclusion, distinguishing between regular and irregular singular points is an important step in the study of differential equations.

Recall Power Series

A power series is an infinite series of the form

Σ a_{n}(x – c)^{n}

where a_{n} is a sequence of constants, x is a variable, and c is a constant. Power series are used to represent functions as an infinite sum of powers of x – c.

Example:

The power series for e^{x} is given by:

Σ x^{n}/n!

where n! denotes the factorial of n. Here, a_{n} = 1/n!, x is the variable, and c = 0.

Properties of Power Series:

- Radius of Convergence: Every power series has a radius of convergence, which is a non-negative real number R or infinity. If the series converges when |x – c| < R, it is said to converge absolutely in that interval.
- Interval of Convergence: The interval of convergence is the set of all values of x for which the power series converges absolutely. The interval may be open, closed, or half-open.
- Continuity: Power series are continuous functions within their interval of convergence.
- Differentiability: If a power series converges to a function f(x), then f(x) is infinitely differentiable on the interval of convergence.

Example:

The power series for sin(x) is given by:

Σ (-1)^{n} x^{(2n+1)}/(2n+1)!

where n! denotes the factorial of n. Here, a_{n} = (-1)^{n}/(2n+1)!, x is the variable, and c = 0.

The radius of convergence for this series is infinity, and the interval of convergence is (-infinity, infinity). The series converges absolutely for all values of x. The function sin(x) is continuous and differentiable on the interval of convergence, and its derivatives can be found by differentiating the power series term-by-term.

Find the Solution of a Differential Equation using Power Series Method

The Power series method is a powerful technique for finding the solution of a differential equation. In this method, we assume that the solution of the differential equation can be represented as a power series. We substitute the power series into the differential equation and equate the coefficients of each power of x. By solving the resulting system of equations, we can determine the coefficients of the power series and thus the solution of the differential equation.

Steps for finding the solution of a differential equation using the power series method:

- Write the differential equation in the form y” + p(x)y’ + q(x)y = 0.
- Assume that the solution of the differential equation can be represented as a power series y(x) = Σ a
_{n}x^{n}. - Find the first few coefficients a
_{0}, a_{1}, a_{2}, etc., by substituting the power series into the differential equation and equating the coefficients of each power of x. - Solve for the remaining coefficients using recurrence relations.
- Write the solution in closed form.

Example:

Find the power series solution of the differential equation y” – xy’ + y = 0.

Solution:

Step 1: Write the differential equation in the form y” + p(x)y’ + q(x)y = 0. Here, p(x) = -x and q(x) = 1.

y” – xy’ + y = 0

Step 2: Assume that the solution of the differential equation can be represented as a power series y(x) = Σ a_{n} x^{n}.

y(x) = Σ a_{n} x^{n}

Step 3: Find the first few coefficients a_{0}, a_{1}, a_{2}, etc., by substituting the power series into the differential equation and equating the coefficients of each power of x.

y” – xy’ + y = 0

Σ n(n-1)a_{n} x^{(n-2)} – xΣ na_{n} x^{(n-1) }+ Σ a_{n} x^{n} = 0

Grouping the terms with the same power of x, we have:

n(n-1)a_{n} – na_{n} + a_{n} = 0

a_{n} = -a_{(n-2)}/(n(n-1)-n)

Therefore, the first few coefficients are:

a_{0} = 1

a_{1} = 0

a_{2 =} -1/2

a_{3} = 0

a_{4} = 1/8

a_{5} = 0

Step 4: Solve for the remaining coefficients using recurrence relations. We can use the formula for an to find the remaining coefficients:

a_{6} = -a_{4}/24

a_{7} = 0

a_{8} = a_{2}/720

a_{9} = 0

a_{10} = -a_{0}/40320

Step 5: Write the solution in closed form.

The power series solution of the differential equation is:

y(x) = Σ a_{n} x^{n}

y(x) = 1 – x^{2}/2 + x^{4}/8 – x^{6}/48 + x^{8}/384 – x^{10}/3840 + …

This is the Taylor series for cos(x), so the solution to the differential equation is y(x) = cos(x).

Recall Frobenius Method

The Frobenius method is a technique used to solve a second-order linear differential equation of the form

x^{2}y” + pxy’ + qy = 0

where p(x) and q(x) are analytic functions at x = 0. This method is particularly useful for finding solutions of differential equations with a regular singular point at x = 0, where p(x) and q(x) may have poles, logarithmic or algebraic singularities.

The method involves assuming a solution of the form

y(x) = x^{r}Σ a_{n} x^{n}

where r is a constant, and a_{n} are constants to be determined. Substituting this solution into the differential equation and equating the coefficients of the powers of x, we can obtain a recurrence relation for the coefficients a_{n}. The solution can then be expressed as a power series that converges in a neighborhood of the regular singular point.

The Frobenius method can produce two linearly independent solutions, and their linear combination gives the general solution of the differential equation.

The method can be applied to a wide range of differential equations in mathematical physics, including the Bessel equation, Legendre equation, and hypergeometric equation.

Example:

Find the Frobenius series solution of the differential equation

x^{2}y” – xy’ + (1 + x)y = 0

Solution:

Step 1: Check if the differential equation has a regular singular point at x = 0. Here, p(x) = -1/x and q(x) = (1 + x)/x^{2}. Both functions are analytic at x = 0. Therefore, x = 0 is a regular singular point.

Step 2: Assume a solution of the form

y(x) = x^{r}Σ a_{n} x^{n}

Substitute the solution into the differential equation:

x^{2}y” – xy’ + (1 + x)y = 0

x^{2}r(r-1)Σ a_{n} x^{(n+r-2)} – xrΣ a_{n} x^{(n+r-1)} + Σ a_{n} x^{(n+r)} + Σ a_{n} x^{(n+r+1)} = 0

Group the terms with the same power of x:

Σ a_{n} x^{(n+r-2)}(n+r)(n+r-1) – Σ a_{n} x^{(n+r-1)}(n+r) + Σ a_{n} x^{(n+r)}(1 + x) + Σ a_{n} x^{(n+r+1)} = 0

Step 3: Equate the coefficients of each power of x to obtain a recurrence relation for the coefficients an .

For n = 0, we have

a_{0}(r(r-1)) = 0

Since x = 0 is a regular singular point, r cannot be a non-negative integer. Therefore, a_{0} = 0.

For n = 1, we have

a_{1}(r(r+1) – 1) = 0

If a_{1} = 0, then the solution is identically zero. This is not a desirable solution, so we assume that a_{1} is nonzero. This gives the indicial equation:

r(r+1) – 1 = 0

Solving the equation, we get

r_{1} = 1/2

r_{2} = -1

Find the Solution of a Differential Equation using Frobenius series method about Regular Singular Point

The Frobenius series method is a powerful technique for solving differential equations with a regular singular point. A regular singular point is a point at which the coefficients of a differential equation are singular but can be expressed as a power series in x. For such points, the Frobenius series method is particularly useful.

The method involves assuming a solution of the form

y(x) = x^{r}Σ a_{n} x^{n}

where r is a constant, and a_{n} are constants to be determined. Substituting this solution into the differential equation and equating the coefficients of the powers of x, we can obtain a recurrence relation for the coefficients a_{n.} The solution can then be expressed as a power series that converges in a neighbourhood of the regular singular point.

The Frobenius series method can produce two linearly independent solutions, and their linear combination gives the general solution of the differential equation.

Example:

Find the Frobenius series solution of the differential equation

x^{2}y” + 2xy’ + (1 – x)y = 0

about the regular singular point x=0.

Solution:

Step 1: Check if the differential equation has a regular singular point at x = 0. Here, p(x) = 2/x and q(x) = (1 – x)/x^{2}. Both functions are analytic at x = 0. Therefore, x = 0 is a regular singular point.

Step 2: Assume a solution of the form

y(x) = x^{r}Σ a_{n} x^{n}

Substitute the solution into the differential equation:

x^{2}y” + 2xy’ + (1 – x)y = 0

x^{2}r(r-1)Σ a_{n} x^{(n+r-2)} + 2xrΣ a_{n} x^{(n+r-1)} + Σ a_{n} x^{(n+r)} – Σ a_{n} x^{(n+r+1)} = 0

Group the terms with the same power of x:

Σ a_{n} x^{(n+r-2)}(n+r)(n+r-1) + Σ a_{n} x^{(n+r-1)}(2n+2r) + Σ a_{n} x^{(n+r)}(1 – x) – Σ a_{n} x^{(n+r+1)} = 0

Step 3: Equate the coefficients of each power of x to obtain a recurrence relation for the coefficients an .

For n = 0, we have

a_{0}(r(r-1)) + a_{0} = 0

(a_{0}r(r-1) + a_{0}) = 0

Since x = 0 is a regular singular point, r cannot be a non-negative integer. Therefore, a_{0} = 0.

For n = 1, we have

a_{1}(r(r-1) + 2r) + a_{1}(2r+2) = 0

a_{1}(r^{2} + 3r + 2) = 0

The roots of this quadratic equation are r = -2 and r = -1. These are the two possible values of r that make a1 nonzero.

Describe Bessel Equation

The Bessel equation is a second-order ordinary differential equation of the form:

x^{2}y” + xy’ + (x^{2} – n^{2})y = 0

where y is a function of x, and n is a constant.

Bessel Equation is used to describe a wide range of physical phenomena, such as heat transfer, fluid flow, and electromagnetic waves. It is named after Friedrich Bessel, a German mathematician who first introduced these functions in 1817.

The solutions of the Bessel equation are known as Bessel functions, denoted as Jn(x), where n is an integer, and x is a real number. Bessel functions can be classified as either regular or irregular. Regular Bessel functions are defined by Jn(x), where the function is finite at x=0. Irregular Bessel functions are defined by Yn(x), where the function is infinite at x=0.

Examples:

Let’s consider a few examples to illustrate the Bessel equation and Bessel functions.

- Heat transfer: In heat transfer, Bessel functions can be used to solve the heat conduction equation in cylindrical coordinates. The equation can be written as:

∂T/∂t = α∇^{2}T

where T is the temperature, t is time, α is the thermal diffusivity, and ∇^{2} is the Laplacian operator. The solution to this equation involves the Bessel functions, which describe the temperature distribution in a cylindrical object.

- Electromagnetic waves: Bessel functions are also used in the study of electromagnetic waves. The Bessel function Jn(x) describes the radial component of the electric field for a cylindrical waveguide.
- Fluid flow: In fluid flow, Bessel functions can be used to describe the velocity distribution in a pipe. The Bessel equation arises from the Navier-Stokes equation, which describes the motion of fluid.
- Mechanical vibrations: Bessel functions are also used to describe mechanical vibrations in cylindrical structures, such as pipes and rods. The displacement of a cylindrical structure undergoing longitudinal vibrations can be described by the Bessel function.

In conclusion, Bessel Equation is an important mathematical tool used in many areas of physics and engineering to describe a wide range of physical phenomena. The solutions to the Bessel equation are known as Bessel functions, which have many practical applications in fields like heat transfer, fluid flow, and electromagnetic waves.

Find the Series Solution of Bessel Equation

Bessel’s equation is a second-order ordinary differential equation that arises in many applications of mathematics and physics, such as in the study of vibrations, electromagnetic waves, and quantum mechanics. The Bessel equation can be written as:

x^{2} y” + x y’ + (x^{2} – n^{2})y = 0

where y is a function of x, and n is a constant. The Bessel equation has a particular solution called the Bessel function of the first kind, denoted by Jn(x). The Bessel function is an important special function that appears in many mathematical and physical applications.

The series solution of the Bessel equation is given by:

y(x) = x^{n} * a_{k} x^{k}

where a_{k} is a constant coefficient that depends on n and the particular solution Jn(x). The series solution converges for all values of x, and it represents the Bessel function of the first kind, Jn(x).

Example 1:

Let’s consider the Bessel equation with n=1, which is given by:

x^{2} y” + x y’ + (x^{2} – 1)y = 0

The particular solution of this equation is the Bessel function of the first kind, J1(x). We can find the series solution of this equation as follows:

y(x) = x^{1} * a_{k} ^{k}

y'(x) = x^{1} * a_{k} (k+1) x^{k}

y”(x) = x^{1} * a_{k} (k+1)k x^{k}

Substituting these expressions into the Bessel equation, we get:

x^{2} a_{k} (k+1)k x^{k} + x a_{k} (k+1) x^{k} + (x^{2} – 1) a_{k} x^{k} = 0

Expanding the terms and collecting the coefficients of x^{k}, we get:

a_{k} [(k+1)^{2} – 1] x^{(k+2)} + a_{k} (k+1) x^{(k+1)} = 0

Since the series solution holds for all values of x, we can equate the coefficients of x^{(k+2)} and x^{(k+1)} to zero, respectively, to obtain the recurrence relation:

a_{(k+2)} = – a_{k} / [(k+1)(k+2)]

a_{(k+1)} = 0

We also have a_{0} = J1(0) and a_{1} = J1′(0), which are the initial conditions of the series solution. The first few terms of the series solution are:

J1(x) = x * [1 – (1/3!)x^{2} + (1/5!)x^{4} – (1/7!)x^{6} + …]

Example 2:

Let’s consider the Bessel equation with n=2, which is given by:

x^{2} y” + x y’ + (x^{2} – 4)y = 0

The Bessel equation with n=2 is given by:

x^{2} y” + x y’ + (x^{2} – 4)y = 0

We can solve this equation using the method of Frobenius, which involves assuming a solution of the form:

y(x) = x^{r} a_{n} x^{n}

Substituting this into the differential equation, we get:

x^{2} [r(r-1) a_{n} x^{(n+r-2)} + r a_{n} x^{(n+r-1)}] + x [ a_{n} x^{(n+r)}] + (x^{2} – 4) a_{n} x^{(n+r)} = 0

We can simplify this expression by reindexing the summations and factoring out x^{(r-2)}, which yields:

x^{r} [(n+r)(n+r-2)a_{n} x^{n} + (n+r)a_{n} x^{(n+1)}] + x^{r} (1 – 4/x^{2})a_{n} x^{n} = 0

For this expression to be zero, each coefficient of x^{n} must be zero. We can set the coefficient of x^{n} to zero and solve for a_{n} to get:

a_{n} = -(1/[(n+r)(n+r-2)+n+r])(1 – 4/x^{2})a_{(n-1)}

For n=0, we have:

a_{0} = -(1/r)(1 – 4/x^{2})a_{1}

Since a_{-1} is an arbitrary constant, we can set it to 1 to obtain:

a_{0} = -1/r

For n=1, we have:

a_{1} = -(1/(r+1))(1 – 4/x^{2})a_{0}

Substituting a_{0}, we get:

a_{1} = -1/[(r+1)r]

Therefore, the solution to the Bessel equation with n=2 is:

y(x) = x^{2} [-1/r – 1/[(r+1)r] x + -(1/[(n+r)(n+r-2)+n+r])(1 – 4/x^{2}) [-1/r – 1/[(r+1)r] x]^{(n-2)} x^{n}]

where r = 2 is a root of the indicial equation, and we have used the initial conditions y(0) = 0 and y'(0) = 0. We can simplify this expression by grouping the terms involving x^{n} and factorizing the denominators:

y(x) = x^{2} [A_{0} J_{2}(x) + A_{1} Y_{2}(x)]

where J_{2}(x) and Y_{2}(x) are the Bessel functions of the first and second kind of order 2, respectively, and A_{0} and A_{1} are constants given by:

A_{0} = -1/4 and A_{1} = -π/4

Describe Bessel Function

Bessel functions are a family of special functions that arise in many areas of physics and engineering, particularly in problems involving wave propagation, heat transfer, and fluid dynamics. Bessel functions are named after Friedrich Bessel, a German mathematician who first studied them in the 19th century.

There are two types of Bessel functions: the Bessel functions of the first kind, denoted by Jn(x), and the Bessel functions of the second kind, denoted by Yn(x). The Bessel functions of the first kind are the solutions to the Bessel equation, which is a second-order ordinary differential equation with a regular singularity at x=0. The Bessel functions of the second kind are linearly independent solutions to the Bessel equation, but they are not regular at x=0.

The Bessel function of the first kind, Jn(x), is defined as:

Jn(x) = (1/π) [cos(x sinθ – nθ)] dθ

where n is an integer, and x is a real variable. The Bessel function of the first kind is an even or odd function of x, depending on whether n is even or odd, respectively. The Bessel function of the first kind has an infinite number of zeros, which are known as the Bessel zeros. The Bessel zeros are important in many applications, such as in the design of circular apertures and resonant cavities.

The Bessel function of the second kind, Yn(x), is defined as:

Yn(x) = (1/π) [sin(x sinθ – nθ)] dθ

where n is an integer, and x is a real variable. The Bessel function of the second kind is also an even or odd function of x, depending on whether n is even or odd, respectively. The Bessel function of the second kind is not defined for x=0 and has singularities at x<0. The Bessel function of the second kind is also known as the Neumann function or the Weber function.

Bessel functions have many important properties and applications, such as:

- Orthogonality: Bessel functions of different orders are orthogonal on the interval [0, a], where a is a positive constant.
- Recurrence relations: Bessel functions satisfy recurrence relations that can be used to compute the function for higher orders.
- Asymptotic behavior: Bessel functions have well-defined asymptotic behaviors for large and small values of x.
- Spherical Bessel functions: Bessel functions can be extended to the three-dimensional case, where they are known as spherical Bessel functions. Spherical Bessel functions have important applications in quantum mechanics and electrodynamics.

Example:

Let’s consider the Bessel function of the first kind with n=0, which is given by:

J0(x) = (1/π) cos(x sinθ) dθ

We can evaluate this integral using the method of complex variables. Let z = x e^{ (iθ)}, then dz = i x e^{(iθ)} dθ. Substituting this into the integral, we get:

J0(x) = (1/π) Re[e^{(iz)}] dz

Using the geometric series expansion, we have:

J0(x) = (1/π) Re[ (i^{k} / k!) z^{k}] dz

J0(x) = (1/π) ∑

Find the Recurrence Relation of Bessel Functions

Recurrence relations are mathematical relations that express a function of a certain order in terms of a function of a lower or higher order. In the case of Bessel functions, there are several recurrence relations that can be used to compute the Bessel functions of higher orders from the Bessel functions of lower orders.

The most commonly used recurrence relation for Bessel functions is the following:

J_{{n+1}}(x) = (2n/x) J_{n}(x) – J_{{n-1}}(x)

where Jn(x) is the Bessel function of the first kind of order n, and x is a real variable. This recurrence relation allows us to compute the Bessel function of any order n, given the Bessel functions of orders n-1 and n-2.

Similarly, there is a recurrence relation for the Bessel functions of the second kind, Y_{n}(x), which is given by:

Y_{{n+1}}(x) = (2n/x) Y_{n}(x) – Y_{{n-1}}(x)

This recurrence relation allows us to compute the Bessel function of the second kind of any order n, given the Bessel functions of orders n-1 and n-2.

In addition to these two recurrence relations, there are other recurrence relations for Bessel functions that involve the derivatives of the Bessel functions, such as:

J’*{n}(x) = (1/2) [J*{n-1}(x) – J_{{n+1}}(x)]

Y’*{n}(x) = (1/2) [Y*{n-1}(x) – Y_{{n+1}}(x)]

where J^{‘}_{n}(x) and Y_{n}(x) are the first derivatives of the Bessel functions of the first and second kinds, respectively.

These recurrence relations can be used to compute the Bessel functions of higher orders from the Bessel functions of lower orders. For example, we can use the first recurrence relation to compute J_{3}(x) in terms of J_{2}(x) and J_{1}(x):

J_{3}(x) = (2(2)/x) J_{2}(x) – J_{1}(x)

Similarly, we can use the second recurrence relation to compute Y_{3}(x) in terms of Y_{2}(x) and Y_{1}(x):

Y_{3}(x) = (2(2)/x) Y_{2}(x) – Y_{1}(x)

These recurrence relations can be used to compute Bessel functions of any order, which is useful in many applications, such as in the design of circular apertures and resonant cavities.

Example:

Let’s use the recurrence relation for the Bessel function of the first kind to compute J_{3}(x) in terms of J_{2}(x) and J_{1(}x):

J_{3}(x) = (2(2)/x) J_{2}(x) – J_{1}(x)

Substituting n=2, we get:

J_{3}(x) = (4/x) J_{2}(x) – J_{1}(x)

Similarly, we can use the recurrence relation for the Bessel function of the first kind to compute J_{4}(x) in terms of J_{3}(x) and J_{2}(x):

J_{4}(x) = (2(3)/x) J_{3}(x) – J_{2}(x)

Substituting the expression for J_{3}(x) that we just derived, we get:

J_{4}(x) = (2(3)/x) [(4/x) J_{2}(x) – J_{1}(x)]

Recall the Orthogonal Property of Bessel Functions

The orthogonality property of Bessel functions is an important property that allows us to express any function as a linear combination of Bessel functions. The orthogonality property states that if m and n are two different integers, then the integral of the product of two Bessel functions of the same order over a certain interval is zero, i.e.,

Jm(x) J_{n}(x) x dx = 0 if m ≠ n

where J_{m}(x) and J_{n}(x) are Bessel functions of the first kind of orders m and n, respectively, and a is a positive constant.

This property is important because it allows us to express any function f(x) as a linear combination of Bessel functions of the same order. Specifically, we can write:

f(x) = [ f(x) J_{n}(x) x dx]/[ J_{n}(x)^{2} x dx] J_{n}(x)

where the sum is taken over all possible values of n, and J_{n}(x) is the Bessel function of the first kind of order n.

The orthogonality property also applies to the Bessel functions of the second kind, Y_{n}(x), which are orthogonal to the Bessel functions of the first kind, J_{n}(x), with respect to the same interval and weight function.

The orthogonality property of Bessel functions is useful in many applications, such as in the analysis of electromagnetic wave propagation in cylindrical waveguides and in solving boundary value problems in mathematical physics.

Example:

Let’s compute the integral of the product of two Bessel functions of different orders to demonstrate the orthogonality property.

We want to find the integral of J_{1}(x)J_{3}(x) over the interval [0,a]:

J_{1}(x) J_{3}(x) x dx

Using the identity for the product of two Bessel functions of different orders, we can write:

J_{1}(x) J_{3}(x) = (1/2)[J_{2}(x) – J_{0}(x)]

Substituting this expression into the integral, we get:

J_{1}(x) J_{3}(x) x dx = (1/2) [ J_{2}(x) x dx – J_{0}(x) x dx]

By the orthogonality property of Bessel functions, the integral of J_{2}(x) over the interval [0,a] is zero, since 2 is not equal to 1. Similarly, the integral of J_{0}(x) over the interval [0,a] is also zero, since 0 is not equal to 1. Therefore, we have:

J_{1}(x) J_{3}(x) x dx = 0

This result shows that the product of two Bessel functions of different orders is orthogonal over the interval [0,a].

Describe the Generating Function of Bessel Function

The generating function of Bessel functions is a powerful tool for deriving various properties of Bessel functions. The generating function is defined as:

G(z, t) = e^{[(z/2)(t – 1/t)]}

where z is a variable and t is a parameter. This function can be used to generate the power series expansion of Bessel functions of the first kind, J_{n}(x), as follows:

J_{n}(x) = (1/n!) (x/2)^{n} ∂ⁿ/∂zⁿ [G(z, t)]|_{(z=0)}

where ∂ⁿ/∂zⁿ denotes the nth derivative with respect to z.

Using the power series expansion of the exponential function, we can rewrite the generating function as:

G(z, t) = J_{n}(x) (z/2)^{n} tⁿ

where J_{n}(x) is the Bessel function of the first kind of order n.

This expression is known as the power series expansion of the generating function. It is useful for deriving various properties of Bessel functions, such as their recurrence relations and their orthogonality properties.

In addition to the generating function for the Bessel functions of the first kind, there are also generating functions for the Bessel functions of the second kind, Y_{n}(x), and for the modified Bessel functions, I_{n}(x) and K_{n}(x).

The generating function of the Bessel functions of the second kind, Y_{n}(x), is given by:

F(z, t) = e^{[(z/2)(t + 1/t)]}

Using a similar approach as for the Bessel functions of the first kind, we can write the power series expansion of the generating function as:

F(z, t) = Y_{n}(x) (z/2)^{n} tⁿ

where Y_{n}(x) is the Bessel function of the second kind of order n.

The generating function for the modified Bessel functions, I_{n}(x) and K_{n}(x), can be written as:

H(z, t) = e^{[(z/2)(t – 1/t)]}

Using the power series expansion of the exponential function, we can write the power series expansion of the generating function as:

H(z, t) = I_{n}(x) (z/2)^{n} tⁿ

or

H(z, t) = K_{n}(x) (z/2)^{n} tⁿ

depending on whether we are considering the modified Bessel function of the first kind or the modified Bessel function of the second kind.

Example:

Let’s use the generating function to derive a recurrence relation for the Bessel functions of the first kind.

Starting with the power series expansion of the generating function, we have:

G(z, t) = J_{n}(x) (z/2)^{n} tⁿ

Taking the nth derivative of the generating function with respect to z, we get:

∂ⁿ/∂zⁿ [G(z, t)] = J_{k}(x) (k choose n) (z/2)^{(k-n)} tⁿ

where (k choose n) denotes the binomial coefficient.

Recall Bessel’s Integral

Bessel’s integral is an important tool for computing Bessel functions of the first kind. It is defined as:

J_{n}(x) = (1/π) (cosnθ – x sin θ) dθ

where J_{n}(x) is the Bessel function of the first kind of order n and x is a real parameter.

This integral can be used to compute Bessel functions of the first kind for any real value of x and any non-negative integer value of n.

To see why this integral works, consider the Fourier series expansion of the function cos(x sin θ). This function can be expanded as:

cos(x sin θ) = J_{n}(x) cos(nθ)

where J_{n}(x) is the Bessel function of the first kind of order n.

This Fourier series expansion can be used to compute the integral for J_{n}(x) by integrating both sides of the equation over the interval [0, π] and using the orthogonality properties of the cosine function:

J_{n}(x) = (1/π) cos(x sin θ) cos(nθ) dθ

= (1/π) cos(nθ) [ J_{m}(x) cos(mθ)] dθ

= (1/π) J_{m}(x) cos(nθ) cos(mθ) dθ

Using the orthogonality relation for the cosine function, we get:

J_{n}(x) = (1/π) cos(nθ – x sin θ) dθ

This is Bessel’s integral, which can be used to compute Bessel functions of the first kind of any order and for any real value of x.

Example:

Let’s use Bessel’s integral to compute J_{1}(x).

Using Bessel’s integral, we have:

J_{1}(x) = (1/π) cos(θ – x sin θ) dθ

To evaluate this integral, we can use the substitution u = θ – x sin θ, which gives du/dθ = 1 – x cos θ.

Using this substitution, we have:

J_{1}(x) = (1/π) cos(u) du/dθ dθ

= (1/π) cos(u) (1 – x cos θ) dθ

= (1/π) cos(u) dθ – (x/π) cos(u) cos θ dθ

The first integral on the right-hand side is equal to zero, since the integral of the cosine function over a full period is zero. The second integral can be evaluated using the identity cos(u) cos θ = (1/2)[cos(u + θ) + cos(u – θ)], which gives:

J_{1}(x) = – (x/π) [cos(u + θ) + cos(u – θ)] dθ

= – (x/π) [sin(u + θ) – sin(u – θ)]|_{(0,π)}

= (2/π) sin x – x J_{0}(x)

where J_{0}(x) is the Bessel function of the first kind of order zero.

Recall Legendre Equation

Legendre Equation is a second-order linear differential equation that arises in the study of the Legendre polynomials. The equation is given by:

(1-x^{2})y” – 2xy’ + n(n+1)y = 0

where y is the unknown function and n is a constant known as the order of the Legendre polynomial. The prime notation denotes differentiation with respect to x.

Legendre Equation is named after Adrien-Marie Legendre who first introduced the Legendre polynomials in his work on celestial mechanics in the 18th century. The equation appears in many areas of physics, including electrostatics, quantum mechanics, and astronomy, and plays an important role in the study of spherical harmonics, which are used to describe the solutions to the Laplace equation on a sphere.

Example 1: In quantum mechanics, the Legendre Equation is used to find the wave functions that describe the behavior of electrons in an atom. The wave functions are solutions to the Schrödinger equation, which is a partial differential equation that can be separated into radial and angular parts. The angular part of the Schrödinger equation is given by the Legendre Equation, with n representing the quantum number that determines the energy and angular momentum of the electron.

Example 2: In astronomy, the Legendre Equation is used to model the gravitational potential of a spherically symmetric object, such as a planet or star. The Legendre polynomials are used to expand the potential in a series, which can then be used to compute the gravitational forces on other objects in the vicinity of the spherically symmetric object.

Overall, the Legendre Equation is a fundamental mathematical tool that appears in many different areas of physics and mathematics. Its solutions, the Legendre polynomials, are widely used in the analysis of spherical and rotational symmetries, and have numerous applications in areas such as signal processing, quantum mechanics, and electrodynamics.

Find the series solution of Legendre Equation

The Legendre Equation is a second-order linear differential equation that arises in the study of Legendre polynomials, which are widely used in many areas of physics and mathematics. The series solution of the Legendre Equation involves finding a power series in x that satisfies the differential equation.

The series solution of the Legendre Equation can be found by assuming that the solution y(x) can be expressed as a power series in x:

y(x) = ∑ a_{n} x^{n}

Substituting this power series into the Legendre Equation, we obtain a recurrence relation for the coefficients a_{n}:

a_{{n+2}} = [(n+2)(n+1) – (n-1)(n)] / [(n+2)(n+3)] a_{n}

This recurrence relation can be used to find the coefficients an recursively, starting from some initial values of a_{0} and a_{1}. The series solution of the Legendre Equation is then given by the power series y(x) = ∑ a_{n} x^{n}.

Example: Let us find the series solution of the Legendre Equation with order n=2.

(1-x^{2})y” – 2xy’ + 6y = 0

Assuming that y(x) can be expressed as a power series in x, we have:

y(x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + …

Differentiating y(x) twice and substituting into the Legendre Equation, we get:

(1-x^{2})[2a_{2} + 6a_{3} x + 12a_{4} x^{2} + …] – 2x[a_{1} + 2a_{2} x + 3a_{3} x^{2} + …] + 6[a_{0} + a_{1} x + a_{2} x^{2} + …] = 0

Simplifying the equation and collecting terms with the same powers of x, we obtain the following recurrence relation for the coefficients an :

a_{{n+2}} = [(n+2)(n+1) – 6] / [(n+2)(n+3)] a_{n}

Starting from a_{0} = 1 and a_{1} = 0, we can recursively calculate the coefficients a_{2}, a_{3}, a_{4}, and so on.

a_{2} = 0

a_{3} = -1/20

a_{4} = 0

a_{5} = 3/560

a_{6} = 0

…

The series solution of the Legendre Equation with n=2 is then given by:

y(x) = 1 – 1/20 x^{2} + 3/560 x^{4} + …

This series solution is a polynomial of degree 2, which is the order of the Legendre polynomial for n=2. The Legendre polynomials with higher orders can also be obtained by finding the series solutions of the Legendre Equation with those orders.

Recall Legendre Polynomial

Legendre Polynomials is a class of orthogonal polynomials that are solutions to the Legendre Equation. These polynomials are named after the French mathematician Adrien-Marie Legendre, who introduced them in the 18th century in his work on celestial mechanics. Legendre Polynomials play a fundamental role in many areas of physics and mathematics, including approximation theory, differential equations, and quantum mechanics.

The Legendre polynomials are defined as the solutions to the Legendre Equation with a fixed order n. The first few Legendre polynomials are given by:

P_{0}(x) = 1

P_{1}(x) = x

P_{2}(x) = (1/2)(3x^{2} – 1)

P_{3(}x) = (1/2)(5x^{3} – 3x)

P_{4}(x) = (1/8)(35x^{4 }– 30x^{2 }+ 3)

The Legendre polynomials have several important properties, including orthogonality, normalisation, and recurrence relations. These properties make them useful in many mathematical and physical applications.

The Legendre polynomials are orthogonal on the interval [-1, 1], with respect to the weight function w(x) = 1. That is, for m ≠ n, the integral of the product of two Legendre polynomials over the interval [-1, 1] is zero:

P_{m}(x) P_{n}(x) dx = 0

if m ≠ n

The Legendre polynomials are also normalised such that:

P_{n}(x)^{2} dx = 2/(2n+1)

The recurrence relation for the Legendre polynomials is given by:

(n+1)P_{{n+1}}(x) = (2n+1)xP_{n}(x) – nP_{{n-1}}(x)

This recurrence relation can be used to compute higher-order Legendre polynomials from lower-order ones. The first few Legendre polynomials can be computed directly from the Legendre Equation, but for higher orders, the recurrence relation is often used.

Example: Let us compute the fourth Legendre polynomial, P_{4}(x), using the recurrence relation.

We can start with the first two Legendre polynomials:

P_{0}(x) = 1

P_{1}(x) = x

Using the recurrence relation, we can compute P_{2}(x):

2xP_{1}(x) – P_{0}(x) = 2x^{2} – 1

P_{2}(x) = (1/2)(3x^{2 }– 1)

We can then compute P_{3}(x):

2xP_{2}(x) – P_{1}(x) = (3/2)x(3x^{2}-1) – x = (3/2)x^{3} – (3/2)x

P_{3}(x) = (1/2)(5x^{3} – 3x)

Finally, we can compute P_{4}(x):

2xP_{3}(x) – P_{2}(x) = 5x(5x^{3} – 3x) – (3x^{2} – 1) = (35/8)x^{4} – (15/4)x^{2} + (3/8)

P_{4}(x) = (1/8)(35x^{4} – 30x^{2} + 3)

Thus, the fourth Legendre polynomial is given by P_{4}(x) = (1/8)(35x^{4} – 30x^{2} + 3).

Recall Rodrigues’ Formula for Legendre Polynomial

Rodrigues’ formula for Legendre polynomials provides an alternative way to express Legendre polynomials as a single expression involving derivatives. Rodrigues’ formula is given by:

P_{n}(x) = (1/2^{n }n!) d^{n}/dx^{n }[(x^{2} – 1)^{n}]

where d^{n}/dx^{n} denotes the nth derivative with respect to x.

This formula provides a convenient way to calculate Legendre polynomials without having to use the recurrence relation or the series solution of the Legendre equation. The formula also demonstrates the fact that Legendre polynomials are polynomials of degree n.

Rodrigues’ formula can be derived by differentiating the power series expansion of the function (x^{2} – 1)^{n}, and then normalizing the resulting polynomial to obtain the Legendre polynomial. The formula can also be used to derive several properties of the Legendre polynomials, including their orthogonality and normalization.

Example: Let us use Rodrigues’ formula to compute the third Legendre polynomial, P_{3}(x).

We start by differentiating the function (x^{2} – 1)^{3} three times:

d/dx [(x^{2} – 1)^{3}] = 6x(x^{2} – 1)^{2}

d^{2}/dx^{2} [(x^{2} – 1)^{3}] = 6(x^{2} – 1)(3x^{2} – 1)

d^{3}/dx^{3} [(x^{2} – 1)^{3}] = 72x(x^{2} – 1)

Substituting these derivatives into Rodrigues’ formula, we get:

P_{3}(x) = (1/2^{3} * 3!) d^{3/}dx^{3 }[(x^{2} – 1)^{3}]

= (1/48) * 72x(x^{2} – 1)

= (1/4) * (5x^{3} – 3x)

Thus, we have shown that the third Legendre polynomial is given by P_{3}(x) = (1/4) * (5x^{3} – 3x), which agrees with the result obtained from the recurrence relation.

Describe Generating Function of Legendre Polynomial

The generating function of Legendre polynomials is a powerful tool used in combinatorics and analysis to generate an infinite sequence of polynomials by means of a single function. The generating function of Legendre polynomials is given by:

G(x, t) = 1 / sqrt(1 – 2tx + t^{2})

where x is the independent variable and t is the parameter. This function generates the sequence of Legendre polynomials P_{n}(x) as the coefficients of the expansion of G(x, t) in powers of t. Specifically, we have:

G(x, t) = ∑ n>=0 P_{n}(x) t^{n}

This means that the nth Legendre polynomial is given by the coefficient of t^{n} in the expansion of G(x, t).

The generating function of Legendre polynomials has several useful properties. For example, it can be used to derive recurrence relations for Legendre polynomials, as well as to prove their orthogonality and completeness. In addition, the generating function can be used to compute integrals involving Legendre polynomials.

Example: Let us use the generating function to compute the fourth Legendre polynomial, P_{4}(x).

We start by expanding the generating function in powers of t:

G(x, t) = 1 / sqrt(1 – 2tx + t^{2})

= ∑ n>=0 t^{n} / sqrt(1 – 2tx + t^{2})

To compute P_{4}(x), we need to find the coefficient of t^{4} in this expansion. We can do this by using the binomial series expansion:

1 / sqrt(1 – 2tx + t^{2}) = ∑ n>=0 (2n+1) / 2 * t^{n} * P_{n}(x)

Substituting this into the expansion of G(x, t), we get:

G(x, t) = ∑ n>=0 t^{n} ∑ k>=0 (2k+1) / 2 * t^{k} * P_{k}(x)

To find the coefficient of t^{4}, we need to look at the terms of the form t^{k} * t^{(4-k)} = t^{4}. The only term that contributes is the one where k=2:

t^{2} * P_{2}(x) * (2(2)+1) / 2 = 3t^{4} * P_{2}(x)

Therefore, the coefficient of t^{4} in the expansion of G(x, t) is 3P_{2}(x). Thus, we have:

P_{4}(x) = [t^{4}] G(x, t) = [t^{4}] (3P_{2}(x) + other terms)

= 0 (since there are no other terms with t^{4})

Therefore, the fourth Legendre polynomial is zero. This can also be verified using the recurrence relation for Legendre polynomials.

Recall Orthogonality of Legendre Polynomial

The orthogonality of Legendre polynomials is a fundamental property of some families of polynomials that allows us to define an inner product on a given function space. Orthogonal polynomials are polynomials that are orthogonal with respect to a given weight function over a certain interval.

The Legendre polynomials are orthogonal with respect to the weight function w(x) = 1 over the interval [-1, 1]. This means that for any two distinct Legendre polynomials P_{n}(x) and P_{m}(x) with n ≠ m, we have:

P_{n}(x) P_{m}(x) dx = 0

In other words, the inner product of P_{n}(x) and P_{m}(x) is zero over the interval [-1, 1].

The orthogonality of Legendre polynomials can be proven using the Rodrigues’ formula, which expresses the Legendre polynomials in terms of derivatives of a function. Specifically, we have:

P_{n}(x) = 1 / 2^{n} * n! * (d^{n} / dx^{n}) [(x^{2} – 1)^{n}]

Using this formula, we can show that the inner product of P_{n}(x) and P_{m}(x) is proportional to a Kronecker delta, which is 1 if n=m and 0 otherwise. The details of this proof are somewhat involved and require techniques from calculus and analysis.

The orthogonality of Legendre polynomials has several important applications in physics and engineering. For example, it is used in the solution of differential equations, numerical analysis, and signal processing. It is also used in the expansion of functions in terms of a series of Legendre polynomials, which is called the Legendre series.

Example: Let us verify the orthogonality of the first few Legendre polynomials.

For n = 0 and m = 1, we have:

P_{0}(x) P_{1}(x) dx = x dx = 0

For n = 1 and m = 2, we have:

P_{1}(x) P_{2}(x) dx = 3x^{2} – 1 dx = 0

For n = 2 and m = 3, we have:

P_{2}(x) P_{3}(x) dx = (5x^{3} – 3x) dx = 0

And so on. In each case, we can use the Rodrigues’ formula to compute the Legendre polynomials and then evaluate the integral using basic calculus. The resulting value is always zero, which confirms the orthogonality of the Legendre polynomials with respect to the weight function w(x) = 1 over the interval [-1, 1].

Recall the Recurrence Relation

The recurrence relation for Legendre polynomials is a mathematical equation that recursively defines a sequence of values. In the case of Legendre polynomials, there is a well-known recurrence relation that can be used to compute the polynomials for higher order values from lower order values.

The recurrence relation for Legendre polynomials is given by:

(n+1)P_{{n+1}}(x) = (2n+1)xPn (x) – nP_{{n-1}}(x)

where P_{n }(x) is the nth Legendre polynomial, and P_{{n-1}}(x) and P_{{n+1}}(x) are the (n-1)th and (n+1)th Legendre polynomials, respectively. This relation provides a way to compute the Legendre polynomials of any order given the values of the lower-order polynomials.

This recurrence relation can be derived from the Rodrigues’ formula for Legendre polynomials, which expresses each polynomial in terms of its derivatives. By using the product rule and some algebraic manipulation, we can obtain the recurrence relation.

The recurrence relation has several applications in physics and engineering, including the solution of differential equations and the computation of integrals. It is also used in the numerical computation of Legendre polynomials, as it provides a fast and efficient way to compute the polynomials of higher order.

Example: Let us use the recurrence relation to compute the first few Legendre polynomials.

We know that P_{0}(x) = 1 and P_{1}(x) = x. Using the recurrence relation, we can compute P_{2}(x) as follows:

2P_{2}(x) = (2*1 + 1)xP_{1}(x) – 1P_{0}(x) = 3x

Therefore, P_{2}(x) = (3/2)x. Using the recurrence relation again, we can compute P_{3}(x) as follows:

3P_{3}(x) = (2*2 + 1)xP_{2}(x) – 2P1(x) = (5x^{2} – 3)/2

Therefore, P_{3}(x) = (5x^{2} – 3)/ (2*3). And so on. By repeatedly applying the recurrence relation, we can compute the Legendre polynomials of any order.

Find the Recurrence Relation of Legendre Polynomial

The recurrence relation of Legendre polynomials is a mathematical equation that recursively defines a sequence of values. In the case of Legendre polynomials, there is a well-known recurrence relation that can be used to compute the polynomials for higher order values from lower order values.

To derive the recurrence relation, we start with the Rodrigues’ formula for Legendre polynomials:

P_{n}(x) = (1/2^{n} n!) d^{n}/dx^{n} [(x^{2} – 1)^{n}]

where P_{n }(x) is the nth Legendre polynomial, and d^{n}/dx^{n} is the nth derivative with respect to x.

We can express the nth derivative of (x^{2} – 1)^{n} in terms of its lower-order derivatives using the product rule:

d^{n}/dx^{n} [(x^{2} – 1)^{n}] = (x^{2} – 1)^{n} d^{n}/dx^{n} [1] + n(x^{2} – 1)^{{n-1}} d^{{n-1}}/dx^{{n-1}} [x^{2}]

The derivative of 1 is 0, so the first term vanishes. The derivative of x^{2 }is 2x, so we can express the second term as:

n(x^{2} – 1)^{{n-1}} d^{{n-1}}/dx^{{n-1}} [^{x2}] = n(x^{2} – 1)^{{n-2}} [(n-1)x^{2} – (n-1)]

Substituting this expression back into the Rodrigues’ formula, we obtain:

P_{n}(x) = (1/2^{n }n!) (2x) d/dx [P_{{n-1}}(x)] – ((n-1)/n) P_{{n-2}}(x)

This is the recurrence relation for Legendre polynomials. It relates the nth polynomial to the (n-1)th and (n-2)th polynomials, and provides a way to compute the Legendre polynomials of any order given the values of the lower-order polynomials.

Example: Let us use the recurrence relation to compute the first few Legendre polynomials.

We know that P_{0}(x) = 1 and P_{1}(x) = x. Using the recurrence relation, we can compute P_{2}(x) as follows:

P_{2}(x) = (1/2^{2} * 2!) (2x) d/dx [P_{1}(x)] – ((1)/2) P_{0}(x) = (3x^{2} – 1)/2

Therefore, P_{2}(x) = (3x^{2} – 1)/2. Using the recurrence relation again, we can compute P_{3}(x) as follows:

P_{3}(x) = (1/2^{3} * 3!) (2x) d/dx [P_{2}(x)] – ((2)/3) P_{1}(x) = (5x^{3} – 3x)/2

Therefore, P_{3}(x) = (5x^{3} – 3x)/2. And so on. By repeatedly applying the recurrence relation, we can compute the Legendre polynomials of any order.