Legendre's equation and Legendre polynomials (intro): Question bank

Setup: every symbol defined, with a picture

The variable $x$ runs on $[- 1, 1]$ because it comes from $x = cos θ$ , where $θ$ is the polar angle on a sphere. The endpoints $x = \pm 1$ are the poles (north and south) of that sphere — look at the mapping figure below.

The numerator carries the whole story of termination: at $m = n$ it becomes $n (n + 1) - n (n + 1) = 0$ , so $a_{n + 2} = 0$ and the series stops, collapsing into a degree- $n$ polynomial. Watch the numerator hit zero in the figure.

The first four polynomials — the "building-block shapes" on the interval — are drawn here so you can see their parity and their shared value $P_{n} (1) = 1$ .

True or false — justify

Legendre's equation has two independent solutions, but only one is a polynomial.

True. It is second-order, so it has two independent solutions;

P_{n}

is the polynomial (terminating) one, while the second,

Q_{n}

, is a non-terminating series that blows up logarithmically at

x = \pm 1

Both solutions of Legendre's equation are bounded on

[- 1, 1]

False. Only

P_{n}

stays bounded at the poles

x = \pm 1

; the second solution

Q_{n}

has a

lo g

singularity there, which is exactly why physics discards it.

The point

x = 0

is a singular point of Legendre's equation.

False. The coefficient of

y^{''}

1 - x^{2}

, which equals

1 \neq = 0

x = 0

, so

x = 0

is an ordinary point — that's precisely why a plain Taylor series works there.

The endpoints

x = \pm 1

are ordinary points because the equation still makes sense there.

False. At

x = \pm 1

the coefficient

1 - x^{2}

vanishes, so they are regular singular points; the equation "makes sense" but solutions can misbehave (one blows up), which is the whole selection story.

For non-integer

n

the series solution still terminates into a polynomial.

False. Termination needs the numerator

m (m + 1) - n (n + 1)

to hit exactly

0

at some integer

m

; that only happens when

n

itself is a non-negative integer, otherwise the series runs forever.

Orthogonality

\int_{- 1}^{1} P_{m} P_{n} d x = 0

for

m \neq = n

is a lucky coincidence of these specific polynomials.

False. It follows structurally from the self-adjoint (Sturm–Liouville) form: different eigenvalues

n (n + 1)

force orthogonal eigenfunctions, as the integration-by-parts argument above shows.

P_{n}

has degree exactly

n

True. The recurrence terminates the series at

m = n

, and the

x^{n}

coefficient is nonzero (leading coefficient

\frac{( 2 n )!}{2 ^{n} ( n ! ) ^{2}}

), so the degree is exactly

n

, not less.

P_{n} (1) = 1

is a theorem you prove from the ODE.

False. It is a normalization convention — a choice pinning the single free constant. The ODE only fixes solutions up to a scalar multiple; we choose

P_{n} (1) = 1

P_{5} (- 1) = - 1

True. Since

P_{n} (- 1) = (- 1)^{n}

and

5

is odd,

P_{5} (- 1) = (- 1)^{5} = - 1

; this also reflects that

P_{5}

is an odd function.

The two arbitrary constants of the general solution correspond to

a_{0}

and

a_{1}

True. The recurrence links coefficients two apart, so all even coefficients grow from

a_{0}

and all odd ones from

a_{1}

— exactly the two free constants a second-order ODE demands.

Spot the error

"The recurrence numerator is

m (m - 1) - n (n + 1)

, since

y^{''}

contributes

m (m - 1)

Error: you dropped the

- 2 x y^{'}

contribution. The

y^{''}

term gives

m (m - 1)

, but

- 2 x y^{'}

adds

- 2 m

; combining,

m (m - 1) - 2 m = m (m + 1) - 2 \cdot (m + 1) + \dots

— cleanly,

m (m - 1) - 2 m = m^{2} - 3 m

, and adding the

n (n + 1)

term the correct grouped numerator is

m (m + 1) - n (n + 1)

. Miss the

- 2 m

and termination at

m = n

breaks.

"Since

P_{n}

is a polynomial, its partner solution must also be a polynomial by symmetry."

Error: symmetry of form doesn't force symmetry of behaviour. The partner

Q_{n}

is the same-parity series that never terminates; it's a genuine non-polynomial with a singularity at the poles.

"We use a Frobenius/indicial analysis at

x = 0

to solve Legendre's equation."

Error: full Frobenius (indicial equation) is for singular points. At

x = 0

, an ordinary point, a plain power series

\sum a_{m} x^{m}

suffices — no indicial roots needed.

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

because that's what the recurrence gives."

Error: the recurrence gives

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

up to a constant. Applying

P_{2} (1) = 1

forces

a_{0} = - \frac{1}{2}

, so

P_{2} (x) = \frac{1}{2} (3 x^{2} - 1)

, not

1 - 3 x^{2}

"Orthogonality means

\int_{- 1}^{1} P_{n}^{2} d x = 0

Error: orthogonality is for distinct indices. The same-index integral is the norm

\int_{- 1}^{1} P_{n}^{2} d x = \frac{2}{2 n + 1} \neq = 0

, which is what makes the expansion coefficients well-defined.

"Rodrigues' formula

\frac{1}{2 ^{n} n !} \frac{d ^{n}}{d x ^{n}} (x^{2} - 1)^{n}

works without the

1/ (2^{n} n!)

Error: dropping that factor still gives a degree-

n

solution of the ODE, but it won't satisfy

P_{n} (1) = 1

. The constant is exactly the normalizer that enforces the convention.

Why questions

Why does the substitution

x = cos θ

appear at all?

The angular part of Laplace's Equation in Spherical Coordinates is naturally in

θ

; setting

x = cos θ

converts it into Legendre's form with

x \in [- 1, 1]

, and the poles

θ = 0, π

become the endpoints

x = \pm 1

Why do we insist on the self-adjoint form when both forms are the same equation?

The self-adjoint (Sturm–Liouville) shape makes the boundary terms in the integration-by-parts vanish at

x = \pm 1

, which is what proves orthogonality via Sturm-Liouville Theory; without that structure you'd never expect the clean

δ_{mn}

integral.

Why does requiring boundedness at the poles force

n

to be a non-negative integer?

Only integer

n

makes the series terminate; for any other

n

the surviving infinite series diverges at

x = \pm 1

. So "bounded at the poles" is the physical rule that selects integers.

Why does the recurrence link coefficients two steps apart instead of one?

The equation contains only parity-preserving pieces on each family —

x^{2}

and

2 x y^{'}

shift parity back,

n (n + 1) y

preserves it — so even and odd powers never mix. Hence

a_{m}

determines

a_{m + 2}

, keeping the two parity families separate.

Why can we expand any nice

f (x)

[- 1, 1]

\sum c_{n} P_{n} (x)

Because the

P_{n}

form a complete orthogonal set (a Sturm–Liouville eigenbasis); orthogonality lets us isolate each

c_{n} = \frac{2 n + 1}{2} \int_{- 1}^{1} f P_{n} d x

, just like Fourier Series uses orthogonal sines and cosines.

Why is the terminating solution called "physical" and the other rejected?

Physical potentials, temperatures, and fields must stay finite everywhere on a sphere, including the poles;

Q_{n}

blows up there, so it cannot describe a real bounded field and is discarded.

Edge cases

For

n = 0

, what does the recurrence do and why is

P_{0} = 1

With

n = 0

the numerator

m (m + 1) - 0 = m (m + 1)

is zero at

m = 0

, killing

a_{2}

immediately, so only

a_{0}

survives — a constant, normalized to

P_{0} (x) = 1

P_{0} (x) = 1

orthogonal to

P_{2} (x) = \frac{1}{2} (3 x^{2} - 1)

, and what does that mean geometrically?

Yes;

\int_{- 1}^{1} 1 \cdot \frac{1}{2} (3 x^{2} - 1) d x = 0

. It means the "flat" pattern and the "ring" pattern carry independent information about a temperature on the sphere.

What happens to the odd-series solution when

n

is even (say

n = 2

For even

n

the even series terminates into

P_{n}

, but the odd series (

a_{1} \neq = 0

) does not terminate — it is the unbounded

Q_{n}

-type solution and is rejected for physics.

Does

P_{n}

ever have degree less than

n

(e.g. leading term cancelling)?

No. The leading coefficient

\frac{( 2 n )!}{2 ^{n} ( n ! ) ^{2}}

is strictly positive for every

n

, so the degree is always exactly

n

; the polynomial can never collapse to lower degree.

At the exact points

x = \pm 1

, does

P_{n}

still satisfy the ODE?

P_{n}

is a polynomial, so it's defined and finite at

\pm 1

, but those are singular points of the equation — the ODE's leading coefficient vanishes there, so it's a boundary of validity, not an interior check.

What is

P_{n} (0)

for odd

n

, and why?

It is $0$ for every odd

n

, because odd

P_{n}

are odd functions (

P_{n} (- x) = (- 1)^{n} P_{n} (x) = - P_{n} (x)

), and any odd function must pass through the origin.

Is a negative degree

n < 0

genuinely new, or does it repeat an existing polynomial?

It repeats: since

n (n + 1)

is symmetric under

n \mapsto - (n + 1)

, degree

n = - 1

gives the same eigenvalue as

n = 0

n = - 2

same as

n = 1

, and so on — so negative integers add nothing beyond

n = 0, 1, 2, \dots

What about half-integer

n

(say

n = \frac{1}{2}

The numerator

m (m + 1) - \frac{1}{2} \cdot \frac{3}{2}

never hits zero for any whole

m

, so the series never terminates and diverges at

x = \pm 1

; there is no bounded polynomial, which is why only non-negative integers are admitted.

n \to \infty

, what happens to the graphs of

P_{n}

[- 1, 1]

They oscillate faster with more zeros (exactly

n

of them in

(- 1, 1)

), yet always stay within

[- 1, 1]

in the interior and pin to

P_{n} (1) = 1

; more oscillation is the polynomial analogue of higher-frequency modes in Fourier Series.