Questions: Legendre Equations and Legendre Polynomials

Question 1

A student applies the Frobenius method to Legendre's equation with n = 1.5 (a non-integer) and finds a convergent power series solution on (−1, 1). Why is this solution problematic for physical applications in spherical geometry?

• A. The series diverges everywhere because the Frobenius method requires integer parameters
• B. A non-integer n produces an infinite series that diverges at x = ±1, making it unsuitable for spherical boundary conditions where solutions must remain bounded at the poles (θ = 0, π)
• C. The recurrence relation cannot be applied when n is non-integer, so no solution can be computed
• D. The solution is valid but cannot be orthogonalized, making expansion impossible

Answer: (B)

The endpoints x = ±1 correspond to the poles of a sphere (cos θ = ±1). Physical solutions must be bounded there. When n is a non-integer, the Frobenius series does not terminate — it continues as an infinite series that diverges at x = ±1. Only when n is a non-negative integer does the recurrence terminate (the coefficient aₙ₊₂ = 0 makes all subsequent terms vanish), producing a polynomial that is automatically bounded everywhere on [−1, 1]. This is why Legendre polynomials, not the general series solutions, appear in physical problems with spherical symmetry.

Question 2

Why are Legendre polynomials useful for expanding arbitrary functions on [−1, 1]?

• A. They are simple low-degree polynomials that approximate any smooth function accurately by Taylor's theorem
• B. They satisfy the orthogonality condition ∫₋₁¹ Pₘ(x)Pₙ(x) dx = 0 for m ≠ n, which allows each expansion coefficient to be determined independently via the inner product without solving a coupled system
• C. They form a complete basis only for polynomial functions, making them useful for polynomial interpolation
• D. Their recurrence relation guarantees convergence of any partial sum to the target function

Answer: (B)

Orthogonality is the property that makes expansions tractable. If you write f(x) = Σ cₙ Pₙ(x) and multiply both sides by Pₘ(x) and integrate, all cross terms ∫ Pₙ Pₘ dx vanish for n ≠ m. Only the n = m term survives, giving an explicit formula: cₘ = [(2m+1)/2] ∫₋₁¹ f(x) Pₘ(x) dx. Each coefficient is determined independently. This is identical in structure to how Fourier coefficients work for trigonometric expansions. Without orthogonality, extracting individual coefficients would require solving an infinite coupled system.

Question 3

Legendre polynomials arise as solutions to Legendre's equation specifically when n is a non-negative integer, because only then does the Frobenius power series terminate to give a polynomial.

• T. True
• F. False

Answer: True

The recurrence relation for coefficients is aₖ₊₂ = −[n(n+1) − k(k+1)] / [(k+2)(k+1)] · aₖ. When n is a non-negative integer, the numerator n(n+1) − n(n+1) = 0 at k = n, making aₙ₊₂ = 0 and all subsequent coefficients vanish. The series terminates after n+1 terms, yielding a polynomial. For non-integer or negative n, no such termination occurs, and the general solution is an infinite series that diverges at the endpoints. The polynomial property is therefore not a convenient feature but a direct consequence of integer n.

Question 4

The recurrence relation (n+1)Pₙ₊₁(x) = (2n+1)x Pₙ(x) − n Pₙ₋₁(x) requires re-deriving each Legendre polynomial from the power series in order to apply it correctly.

• T. True
• F. False

Answer: False

The recurrence relation is precisely the tool that makes re-derivation unnecessary. Given P₀(x) = 1 and P₁(x) = x, you can compute any subsequent Pₙ algebraically: P₂ = (3x² − 1)/2, P₃ = (5x³ − 3x)/2, and so on, by applying the recurrence repeatedly. This is far more efficient than re-running the Frobenius series each time. The recurrence relation is one of Legendre polynomials' most practically useful properties, enabling rapid computation of arbitrarily high-degree polynomials from the initial two.

Question 5

Explain why the termination of the Frobenius power series is critical to Legendre polynomials' usefulness, and what happens when n is not a non-negative integer.

Model answer: When n is a non-negative integer, the recurrence relation forces the coefficient aₙ₊₂ to zero, causing the infinite series to truncate into a polynomial of degree n. Polynomials are bounded everywhere on [−1, 1], including at the endpoints x = ±1 (the poles of a sphere). This boundedness is physically required — solutions to Laplace's equation in spherical coordinates must not diverge at the poles. When n is not a non-negative integer, the series does not terminate, remains an infinite series, and diverges at x = ±1, making it physically inadmissible for spherical problems. Termination is therefore the selection mechanism that picks out the physically meaningful solutions.

The deeper point is that the physical boundary condition (boundedness at the poles) and the mathematical property (series termination at integer n) are not independent — they are two descriptions of the same constraint. The mathematical structure of the Frobenius solution embeds the physics: only when n(n+1) is an eigenvalue of the Legendre operator do bounded solutions exist. This connection between eigenvalue problems and orthogonal polynomial families is the central pattern that recurs throughout mathematical physics (Hermite polynomials for the harmonic oscillator, Laguerre for the hydrogen atom, etc.).