Question

Obtain the recurrence relations for the solution of Legendre's equation (16.35) in inverse powers of \(z\), i.e. set \(y(z)=\sum a_{n} z^{\sigma-n}\), with \(a_{0} \neq 0 .\) Deduce that if \(\ell\) is an integer then the series with \(\sigma=\ell\) will terminate and hence converge for all \(z\) whilst that with \(\sigma=-(\ell+1)\) does not terminate and hence converges only for \(|z|>1\).

Short answer

The series with \(\sigma=\ell\) terminates and converges for all \(z\), while that with \(\sigma=-(\ell+1)\) does not terminate and converges only for \(|z|>1\).

Step-by-step solution

- Substitute the series into Legendre's equation

First, recall Legendre's equation: \[ (1-z^2)y'' - 2zy' + \rho(\rho+1)y = 0 \] Given the series solution: \[ y(z) = \sum a_{n} z^{\sigma - n} \] Substitute this series into Legendre's equation.

- Compute the derivatives

Calculate the first and second derivatives of the series: \[ y'(z) = \sum (\sigma - n) a_{n} z^{\sigma - n - 1} \] \[ y''(z) = \sum (\sigma - n)(\sigma - n - 1) a_{n} z^{\sigma - n - 2} \]

- Substitute the derivatives back into the equation

Substitute the derivatives back into Legendre's equation: \[ (1-z^2) \sum (\sigma - n)(\sigma - n - 1) a_{n} z^{\sigma - n - 2} - 2z \sum (\sigma - n) a_{n} z^{\sigma - n - 1} + \rho(\rho+1)\sum a_n z^{\sigma - n} = 0 \]

- Simplify and equate coefficients

Reorganize the terms, combine like powers of coefficients, and use the fact that the series must satisfy the equation term-by-term: \[ (\sigma - n)(\sigma - n - 1) a_{n} - (\sigma - n+2) (\sigma - n+1) a_{n-2} + \rho(\rho+1) a_{n} = 0 \]

- Obtain the recurrence relation

Solve for the recurrence relation: \[ a_{n} = (\sigma - n + 2)(\sigma - n + 1) a_{n-2}/[(n-\sigma)(n-\sigma - 1) - \rho(\rho+1)] \]

- Analyze the series for different values of \(\sigma\)

Analyze the behavior of the series for different values of \(\sigma\): - For \(\sigma = \ell\), the series will terminate, converging for all z. - For \(\sigma = - (\ell + 1)\), the series does not terminate, converging only for \|z\|>1.

Key concepts

Recurrence Relations

Recurrence relations are equations that recursively define sequences. In other words, they express each term of the sequence as a function of the preceding terms. For Legendre's equation, after substituting the series solution, we derive a recurrence relation by equating coefficients of like powers of z. This relation helps us compute the coefficients of our series solution step by step.

To see how this works: consider the recurrence relation derived in the step-by-step solution:
\[ a_{n} = \frac{(\theta - n+2)(\theta - n+1) a_{n-2}}{(n-\theta)(n-\theta -1) - \rho(\rho +1)} \]
Here, each term \(a_{n}\) is computed using previous terms (in this case, \(a_{n-2}\)).

• Set initial conditions: Select starting values like \(a_0\) and \(a_1\) based on the context of the specific differential equation.
• Iterate: Use the recurrence to find subsequent terms.

Series Solutions

Series solutions involve expressing a function as a sum of terms in a series, often power series. In solving differential equations like Legendre’s, this method is particularly useful. We express the solution \(y(z)\) as a series: \[ y(z) = \sum a_{n} z^{\theta -n} \]

• Write the series: Substitute the series form of \(y\) and its derivatives into the differential equation.
• Find coefficients: Each term's coefficient must equal zero for the entire sum to be zero. This gives us a relation for the \(a_n\) coefficients.

In solving Legendre's equation, by matching coefficients of powers of \(z\), a recurrence relation is found which helps to generate the series terms for the solution. Whether the series is finite or infinite depends on the behavior of these coefficients.

Differential Equations

Differential equations involve functions and their derivatives. They express a relationship between a function and its rates of change. For instance, Legendre's differential equation is:
\[ (1 - z^2)y'' - 2zy' + \rho(\rho + 1)y = 0 \]
It describes how \(y\)'s second derivative and first derivative relate to \(y\) itself, all depending on \(z\).

To solve differential equations using series solutions:

• Assume a form: Here, we assume that \(y\) can be written as a power series.
• Substitute: Plug this series and its derivatives into the differential equation.
• Simplify: Combine like terms and equate the coefficients to zero to form recurrence relations.

Legendre’s equations are common in physics, especially in problems involving spherical coordinates like gravitational and electromagnetic fields.

Convergence Analysis

Convergence analysis determines where (for which values of \(z\)) a series solution is valid. For polynomial series solutions, we check whether the series converges. In Legendre's problem, we have:

• \textbf{Termination}: If \(\theta\) is an integer, the series terminates. This means all coefficients beyond a certain point are zero, making the series finite. Finite series always converge for all \(z\).
• \textbf{Non-termination}: If \(\theta = - (\rho + 1)\), the series doesn’t terminate. Instead, it continues indefinitely and converges only for \(|z| > 1\).

To fully understand how a series behaves, we analyze these properties, ensuring our solution is valid within the needed domain. Convergence can be checked using various tests like the ratio test, ensuring we have a properly defined solution for the problem at hand.