Legendre Polynomials

The previous article did something slightly underhanded. Separating Laplace's equation in spherical coordinates, it solved the radial equation honestly — $A_\ell r^\ell + B_\ell r^{-\ell-1}$ , in full — but on reaching the polar-angle equation it simply announced that the acceptable solutions are "the Legendre polynomials," listed a few, and hurried on. That IOU comes due now: an electromagnetism student minding their own business suddenly needs an entire polynomial family. Here is the claim to be earned: on a sphere, the functions $P_0, P_1, P_2, \dots$ are not one basis among many — they are the angular DNA of every azimuthally symmetric electrostatics problem. Every potential you will compute on or around a sphere is a superposition of these shapes and no others: $\ell=0$ a uniform skin, $\ell=1$ a north–south split, $\ell=2$ a quadrupolar banding, and so on up. They do for the polar angle what sines do for a line: decompose any boundary data, mode by mode, one integral per coefficient. Mastering them buys three things. Immediate power: the hemisphere and sphere-in-a-field problems of this chapter reduce to reading off coefficients. A transferable skill: the termination argument quantizing $\ell(\ell+1)$ here later quantizes angular momentum in quantum mechanics. And a secret weapon: the generating function proved at the end is the multipole expansion, three articles early.

Key takeaways

Legendre polynomials $P_\ell(x)$ , $x=\cos\theta$ , are the solutions of Legendre's equation finite at the poles $x=\pm1$ ; non-terminating series diverge there, so finiteness forces the separation constant to $\ell(\ell+1)$ , integer $\ell\ge0$ — a physical boundary condition, not a convention.
Rodrigues' formula $P_\ell(x)=\frac{1}{2^\ell \ell!}\frac{\mathrm{d}^\ell}{\mathrm{d}x^\ell}(x^2-1)^\ell$ generates them all, with parity $(-1)^\ell$ and normalization $P_\ell(1)=1$ .
Orthogonality, $\int_{-1}^{1}P_\ell P_{\ell'}\,\mathrm{d}x=\frac{2}{2\ell+1}\delta_{\ell\ell'}$ , makes coefficients a one-integral job: $c_\ell=\frac{2\ell+1}{2}\int_{-1}^{1}f\,P_\ell\,\mathrm{d}x$ .
The generating function $\left(1-2xt+t^2\right)^{-1/2}=\sum_\ell P_\ell(x)\,t^\ell$ is the Coulomb kernel in disguise — the engine the Multipole Expansion article runs on.
Key idea: on a sphere with azimuthal symmetry, the polar-angle modes that stay finite at the poles are exactly the Legendre polynomials, and the full potential is $\sum_\ell(A_\ell r^\ell+B_\ell r^{-\ell-1})P_\ell(\cos\theta)$ with coefficients fixed by orthogonality against the boundary data.

Where this lives today

Every finite-element and spectral code in production — the solvers behind aircraft electromagnetics, MRI coils, weather models — evaluates its integrals by Gauss–Legendre quadrature: sample the integrand at the $n$ roots of $P_n(x)$ , weight appropriately, and the result is exact for polynomials up to degree $2n-1$ . Three points buy exactness through degree five (nodes $0$ and $\pm\sqrt{3/5}\approx\pm0.7746$ , weights $\tfrac89$ and $\tfrac59$ , integrating $x^4$ to $\tfrac25$ on the nose) — this article's orthogonality theorem in industrial dress: the roots of $P_n$ are the sampling points against which lower polynomials cannot hide. Billions of them run every second in large simulations. The original gravitational application never went away. Earth's gravity is a Legendre ladder whose $\ell=2$ coefficient — $J_2=1.08263\times10^{-3}$ multiplying $P_2(\cos\theta)$ , the equatorial bulge — dominates every satellite orbit's perturbations; sun-synchronous missions deliberately surf its precession. Modern geopotential models push the series to degree 2190 — Legendre's 1782 idea, executed to the two-thousandth rung.

The physical picture

Meet the family first, formalities later — the first five members against $x=\cos\theta$ , south pole ( $x=-1$ ) to north pole ( $x=+1$ ):

Figure

The first five Legendre polynomials — every curve arrives at the north pole with value 1

Read the gallery like the modes of a guitar string: $P_0=1$ never crosses zero, $P_1=x$ crosses once (at the equator), $P_2$ twice, $P_3$ three times — the degree $\ell$ counts sign changes as the harmonic number counts nodes. And every curve arrives at the north pole with value $1$ : the convention $P_\ell(1)=1$ , which lets every mode report its coefficient undisguised on the axis. Now wrap each curve around a sphere. Since $x=\cos\theta$ depends only on latitude, each $P_\ell(\cos\theta)$ paints the sphere in latitude bands of alternating sign — the charge-pattern vocabulary of spherical electrostatics. A surface charge $\sigma \propto P_1$ is the north-positive, south-negative skin a uniform field induces on a conductor; a $\sigma \propto P_2$ skin has charged polar caps and an opposite tropical belt. Any axisymmetric pattern is a mix of these bands.

Figure

Charge-pattern vocabulary on a sphere — the latitude bands painted by the first three modes

Two questions should be nagging you: why polynomials, and why the strange eigenvalues $\ell(\ell+1) = 0, 2, 6, 12, \dots$ ? Both have one answer, and it is physical: the poles of the sphere are ordinary points of space, and the potential there must be finite. Everything else here is machinery built on that single quantization.

The equation and the toolkit

The angular equation is inherited, not derived: as we found in Separation of Variables in Spherical Coordinates, separating $V=R(r)\Theta(\theta)$ with azimuthal symmetry hands the polar angle its own ODE, its separation constant written suggestively as $\ell(\ell+1)$ but not yet restricted. The substitution $x=\cos\theta$ (so $\mathrm{d}x=-\sin\theta\,\mathrm{d}\theta$ , $\sin^2\theta=1-x^2$ ) converts it to standard form:

\frac{\mathrm{d}}{\mathrm{d}x}\!\left[(1-x^2)\frac{\mathrm{d}P}{\mathrm{d}x}\right]+\ell(\ell+1)\,P=0 \qquad (1)

where: $x=\cos\theta$ runs over $[-1,1]$ pole to pole; $P(x)$ — the angular profile we seek; $\ell(\ell+1)$ — the separation constant, at this stage any number. This is Legendre's differential equation, and it carries a structural warning: the coefficient $(1-x^2)$ of the highest derivative vanishes at $x=\pm1$ — singular points sitting exactly at the poles, where solutions go to die. The first derivation prices survival there at quantization. One compact machine generates every member — Rodrigues' formula:

P_\ell(x)=\frac{1}{2^\ell\,\ell!}\,\frac{\mathrm{d}^\ell}{\mathrm{d}x^\ell}\big(x^2-1\big)^\ell \qquad (2)

where: $(x^2-1)^\ell$ — a degree- $2\ell$ polynomial; $\ell$ -fold differentiation leaves degree $\ell$ ; the prefactor enforces $P_\ell(1)=1$ . Verify on low orders: $\ell=1$ gives $\frac12\frac{\mathrm{d}}{\mathrm{d}x}(x^2-1)=x$ ✓; $\ell=2$ gives $\frac18(12x^2-4)=\frac12(3x^2-1)$ ✓. Parity comes free: $(x^2-1)^\ell$ is even and each derivative flips parity once, so $P_\ell(-x)=(-1)^\ell P_\ell(x)$ , whence $P_\ell(-1)=(-1)^\ell$ . Running the machine — or the recurrence $(\ell+1)P_{\ell+1}=(2\ell+1)x\,P_\ell-\ell\,P_{\ell-1}$ , seeded by $P_0=1$ , $P_1=x$ — produces the working table:

$\ell$	$P_\ell(x)$	parity	$P_\ell(0)$
0	$1$	even	$1$
1	$x$	odd	$0$
2	$\tfrac{1}{2}(3x^2-1)$	even	$-\tfrac12$
3	$\tfrac{1}{2}(5x^3-3x)$	odd	$0$
4	$\tfrac{1}{8}(35x^4-30x^2+3)$	even	$\tfrac38$
5	$\tfrac{1}{8}(63x^5-70x^3+15x)$	odd	$0$

Orthogonality turns the table into a solving method: distinct members are perpendicular on $[-1,1]$ , with known "length squared":

\int_{-1}^{1}P_\ell(x)\,P_{\ell'}(x)\,\mathrm{d}x=\frac{2}{2\ell+1}\,\delta_{\ell\ell'} \qquad (3)

where: $\delta_{\ell\ell'}$ — the Kronecker delta; $2/(2\ell+1)$ — the normalization, shrinking as faster-wiggling polynomials cancel more of themselves. In $\theta$ language the same integral reads $\int_0^\pi P_\ell P_{\ell'}\sin\theta\,\mathrm{d}\theta$ — the $\sin\theta$ is $|\mathrm{d}x|$ , not decoration, and forgetting it corrupts every coefficient. The second derivation proves (3) in full from Rodrigues' formula. Orthogonality plus completeness — the $P_\ell$ span everything square-integrable on $[-1,1]$ — yields the Legendre series. Any reasonable $f(x)$ expands as $f=\sum_\ell c_\ell P_\ell$ , and Fourier's trick (multiply by $P_{\ell'}$ , integrate; orthogonality annihilates all terms but one) extracts each coefficient:

c_\ell=\frac{2\ell+1}{2}\int_{-1}^{1}f(x)\,P_\ell(x)\,\mathrm{d}x \qquad (4)

where: $\frac{2\ell+1}{2}$ — the inverse of the normalization in (3); $f$ — your boundary data, typically a potential $V_0(\theta)$ read as a function of $x=\cos\theta$ . It is the exact analogue of a Fourier coefficient — $P_\ell$ in the role of the Cartesian article's $\sin(n\pi y/a)$ . Parity gives instant shortcuts: even data excite only even $\ell$ ; odd data only odd. And once the series matches the boundary, the uniqueness theorems certify it as the answer. The last tool is the deepest. The whole family is packaged inside one closed form, the generating function:

\frac{1}{\sqrt{1-2xt+t^2}}=\sum_{\ell=0}^{\infty}P_\ell(x)\,t^{\ell}, \qquad |t|<1 \qquad (5)

where: $t$ — a bookkeeping parameter whose powers sort the family by degree; $x$ — the usual cosine. Expand in powers of $t$ and the coefficient of $t^\ell$ is, every time, exactly $P_\ell(x)$ — a party trick, until you recognize the left-hand side as the Coulomb potential of a point charge expanded in a distance ratio. The third derivation proves it — mark the result well: this identity is the engine on which the Multipole Expansion of the Potential article, three articles ahead, will run.

Derivation ASeries solution and the quantization of the separation constant

The goal is the article's central claim: Legendre's equation has solutions finite at both poles only when the separation constant equals $\ell(\ell+1)$ for a non-negative integer $\ell$ , and those solutions are polynomials. The strategy is the power-series (Frobenius) method — assume a series, convert the equation into a recursion among coefficients, then interrogate it about the endpoints. The step to watch is the honest part: showing a non-terminating series actually diverges at $x=\pm1$ .

Step 1. Expand the equation to $(1-x^2)P''-2xP'+\lambda P=0$ , $\lambda$ unrestricted, and substitute $P(x)=\sum_{k}a_k x^k$ . Collecting the coefficient of $x^k$ (the $(1-x^2)P''$ term contributes at two offsets) and using $k(k-1)+2k=k(k+1)$ :

a_{k+2}=\frac{k(k+1)-\lambda}{(k+1)(k+2)}\,a_k

The recursion links $k$ to $k+2$ : even coefficients descend from $a_0$ , odd from $a_1$ — two independent solutions, as a second-order equation must have.

Step 2. Suppose the series never terminates. For large $k$ , $a_{k+2}/a_k \to k/(k+2)\approx 1-2/k$ — the coefficient pattern of logarithm-type series: $a_{2m}\sim C/m$ , so at $x=\pm1$ the tail is the harmonic series $\sum_m 1/m$ . The solution diverges logarithmically at both endpoints. Why this is fatal: $x=\pm1$ is the polar axis — ordinary empty space, nothing singular about the physics there. A potential that blows up on the axis is an artifact to discard. (The divergent partners survive as the second-kind functions $Q_\ell$ — legitimate in problems that exclude the axis, such as fields between cones.)

Step 3. The only escape is termination. If $\lambda=\ell(\ell+1)$ for a non-negative integer $\ell$ , the numerator vanishes at $k=\ell$ , so $a_{\ell+2}=a_{\ell+4}=\dots=0$ : one series becomes a polynomial of degree $\ell$ . Kill the other ( $a_1=0$ for even $\ell$ , $a_0=0$ for odd) and the survivor is finite on all of $[-1,1]$ . No other $\lambda$ terminates either series: the separation constant is quantized by the geometry of the sphere.

Step 4. Normalize by $P_\ell(1)=1$ . Check $\ell=2$ : $\lambda=6$ gives $a_2=-3a_0$ , $a_4=0$ , so $P\propto a_0(1-3x^2)$ ; the value $1$ at $x=1$ fixes $a_0=-\tfrac12$ , i.e. $P_2=\tfrac12(3x^2-1)$ , matching the table. Verified: the recursion, run symbolically for $\ell=0\ldots7$ , matches Rodrigues' formula at every order; the non-terminating $\lambda=1$ even series grows without bound as $x\to1$ numerically.

Go deeper · Derivation B · 1 minOrthogonality and normalization from Rodrigues' formula

Derivation BOrthogonality and normalization from Rodrigues' formula

The goal is equation (3) in both halves: distinct- $\ell$ integrals vanish, and the diagonal equals $2/(2\ell+1)$ . Of the two classic routes, Sturm–Liouville self-adjointness gets the zeros elegantly but must borrow the normalization; Rodrigues proves both in one sustained calculation, so we take it. The strategy: repeated integration by parts — Rodrigues writes $P_\ell$ as an $\ell$ -fold derivative, and each round peels one derivative off. Watch why every boundary term dies: each still contains undifferentiated factors of $(x^2-1)$ , which vanish at both endpoints.

Step 1. Take $\ell' \le \ell$ and insert Rodrigues once:

I=\int_{-1}^{1}P_{\ell'}P_\ell\,\mathrm{d}x=\frac{1}{2^\ell \ell!}\int_{-1}^{1}P_{\ell'}(x)\,\frac{\mathrm{d}^\ell}{\mathrm{d}x^\ell}(x^2-1)^\ell\,\mathrm{d}x

Step 2. Integrate by parts $\ell$ times, each round moving one derivative onto $P_{\ell'}$ . Every boundary term contains a derivative of $(x^2-1)^\ell$ of order below $\ell$ ; since $(x^2-1)^\ell$ has zeros of order $\ell$ at both endpoints, all boundary terms vanish. After $\ell$ rounds:

I=\frac{(-1)^\ell}{2^\ell \ell!}\int_{-1}^{1}(x^2-1)^\ell\,\frac{\mathrm{d}^\ell P_{\ell'}}{\mathrm{d}x^\ell}\,\mathrm{d}x

Step 3. Read off the off-diagonal case: if $\ell'<\ell$ , the $\ell$ -th derivative of the degree- $\ell'$ polynomial $P_{\ell'}$ is identically zero, so $I=0$ . Orthogonality proved — a low-degree polynomial cannot survive that many differentiations. Why this had to happen structurally: Legendre's equation is self-adjoint with the boundary factor $(1-x^2)$ vanishing at the ends, so eigenfunctions with distinct eigenvalues are automatically orthogonal — the Rodrigues computation is Sturm–Liouville theory caught in the act.

Step 4. The diagonal, $\ell'=\ell$ : only the leading term of $P_\ell$ survives $\ell$ differentiations. Rodrigues gives leading coefficient $\frac{(2\ell)!}{2^\ell(\ell!)^2}$ , so $\frac{\mathrm{d}^\ell P_\ell}{\mathrm{d}x^\ell}=\frac{(2\ell)!}{2^\ell \ell!}$ , a constant. With $(x^2-1)^\ell=(-1)^\ell(1-x^2)^\ell$ absorbing the sign:

I=\frac{(2\ell)!}{2^{2\ell}(\ell!)^2}\int_{-1}^{1}(1-x^2)^\ell\,\mathrm{d}x

Step 5. The remaining integral is a Wallis reduction ( $x=\cos\phi$ ): $\int_{-1}^{1}(1-x^2)^\ell\mathrm{d}x=\frac{2^{2\ell+1}(\ell!)^2}{(2\ell+1)!}$ — check $\ell=2$ : direct integration gives $\tfrac{16}{15}$ , the formula $\tfrac{2^5\cdot4}{120}=\tfrac{16}{15}$ ✓. Multiplying, the factorial towers collapse:

I=\frac{(2\ell)!}{2^{2\ell}(\ell!)^2}\cdot\frac{2^{2\ell+1}(\ell!)^2}{(2\ell+1)!}=\frac{2\,(2\ell)!}{(2\ell+1)!}=\frac{2}{2\ell+1}

Verified: diagonal integrals computed symbolically for $\ell=0\ldots5$ give $2,\ \tfrac23,\ \tfrac25,\ \tfrac27,\ \tfrac29,\ \tfrac{2}{11}$ — exactly $2/(2\ell+1)$ — and the cross integrals $(0,2),(1,3),(2,4)$ all return zero; matches Jackson Eq. 3.21.

The generating function — the multipole engine

Go deeper · Derivation C · 2 minThe generating function from the potential of a point charge

Derivation CThe generating function from the potential of a point charge

The goal is identity (5): the expansion coefficients of $g(x,t)=(1-2xt+t^2)^{-1/2}$ in powers of $t$ are exactly the Legendre polynomials. Binomial expansion can verify finitely many orders but proves nothing about all of them; instead we let electrostatics do the work. The strategy: recognize $g/r$ as the potential of a point charge, guaranteed harmonic away from the source; expand in powers of $t=r'/r$ ; show harmonicity forces each coefficient function to satisfy Legendre's equation, while pole-finiteness and one check at $x=1$ pin the coefficients exactly. Watch the family's three defining properties — the ODE, finiteness, $P_\ell(1)=1$ — each get consumed once.

Step 1. Let $u=1/|\mathbf{r}-\mathbf{r}'|$ with the source on the $z$ -axis at radius $r'$ . Away from the charge $u$ is a point-charge potential, so $\nabla^2u=0$ there — the physics input. By the law of cosines, for $r>r'$ :

u=\frac{1}{r}\,g\!\left(\cos\theta,\ \frac{r'}{r}\right), \qquad g(x,t)=\frac{1}{\sqrt{1-2xt+t^2}}

Step 2. Expand $g$ via the binomial series of $(1-w)^{-1/2}$ in $w=2xt-t^2$ (convergent for $|t|<1$ ); collecting powers of $t$ yields some coefficient functions $K_\ell(x)$ , each a polynomial of degree $\ell$ :

u=\sum_{\ell=0}^{\infty}K_\ell(x)\,\frac{r'^{\,\ell}}{r^{\ell+1}}

Step 3. Impose harmonicity term by term. Since $\nabla^2u=0$ for every admissible $r'$ and the terms carry independent powers $r'^{\,\ell}$ , each term must be harmonic on its own. But $f(\cos\theta)/r^{\ell+1}$ is harmonic precisely when $f$ satisfies Legendre's equation (1) with eigenvalue $\ell(\ell+1)$ — the radial-angular bookkeeping of the spherical separation article run in reverse. Why this pins the family: by the first derivation, pole-finite solutions at eigenvalue $\ell(\ell+1)$ form a one-dimensional set — multiples of $P_\ell$ . Our $K_\ell$ is a polynomial, certainly finite at the poles, so $K_\ell=c_\ell P_\ell$ for some constant.

Step 4. Fix the constants at $x=1$ (field point on the axis beyond the charge), where the closed form collapses to a geometric series:

g(1,t)=\frac{1}{\sqrt{1-2t+t^2}}=\frac{1}{1-t}=\sum_{\ell=0}^{\infty}t^\ell

so every coefficient equals $1$ there: $c_\ell P_\ell(1)=1$ , and the convention $P_\ell(1)=1$ forces $c_\ell=1$ for all $\ell$ . Identity (5) is proved, all orders at once.

Step 5. Cross-examine by direct expansion: the binomial series to fourth order gives $1+xt+\tfrac12(3x^2-1)t^2+\tfrac12(5x^3-3x)t^3+\tfrac18(35x^4-30x^2+3)t^4$ — the table's $P_0\ldots P_4$ on the nose. Numerically, at $x=0.6$ , $t=0.3$ : the closed form gives $1.170411$ ; partial sums through $t^3,t^4,t^5$ give $1.173880$ , $1.170575$ , $1.170204$ — converging geometrically onto the exact value. Verified: symbolic expansion to order $t^4$ matches the table; a sixty-term numeric sum at $(x,t)=(0.6,0.3)$ agrees with the closed form to machine precision; the $x=1$ check is exact. Proved in full deliberately: the Multipole Expansion of the Potential article will recall this identity by name, substitute $t=r'/r$ , and never re-derive it.

Limiting cases

case	result	meaning
the poles	$P_\ell(1)=1$ , $P_\ell(-1)=(-1)^\ell$	normalization + parity; on-axis every mode reports its coefficient
equator	$P_{\text{odd}}(0)=0$ ; $P_2(0)=-\tfrac12$ , $P_4(0)=\tfrac38$	odd modes have an equatorial node; even values alternate and shrink
large $\ell$	$P_\ell(\cos\theta)\approx\sqrt{\tfrac{2}{\pi \ell\sin\theta}}\,\cos\!\big[(\ell+\tfrac12)\theta-\tfrac{\pi}{4}\big]$	high modes oscillate like cosines ( $\ell=50$ , $\theta=60°$ : exact $-0.0311$ , asymptotic $-0.0314$ )
smooth data	$c_\ell\to0$ faster than any power	degree- $n$ data needs only $\ell\le n$ : see $x^2$ below
discontinuous data	$c_\ell\sim\ell^{-3/2}$	slow, oscillatory convergence on the sphere — see the hemispheres

Common trapMy boundary data involves sine of theta, so Legendre polynomials in cos theta cannot be enough

Tempting, and exactly backwards. The $P_\ell(\cos\theta)$ are a complete set on $-1\le x\le1$ : anything square-integrable — including $\sin\theta=\sqrt{1-x^2}$ , cusps, even step functions — expands in them, with no second "sine-like" family needed. (Fourier needs two families because its interval is a periodic seam; this one is not.) The companion trap is the weight: orthogonality lives in $x$ , so in $\theta$ language every coefficient integral carries $\sin\theta\,\mathrm{d}\theta$ , and the normalization is $2/(2\ell+1)$ — not $1$ . Coefficients off by factors like $\tfrac23$ or $\tfrac25$ mean a dropped weight or a dropped $\tfrac{2\ell+1}{2}$ .

Worked examples

1 · Expanding a plain polynomial (confidence builder)

Expand $f(x)=x^2$ in Legendre polynomials and confirm the coefficients by formula (4).

Step 1. Degree logic first: degree 2 means only $\ell\le2$ appear; parity (even) kills $\ell=1$ . Write $x^2=c_0P_0+c_2P_2$ and match powers: $\tfrac{3c_2}{2}=1$ gives $c_2=\tfrac23$ ; the constant term $c_0-\tfrac{c_2}{2}=0$ gives $c_0=\tfrac13$ :

x^2=\tfrac{1}{3}P_0(x)+\tfrac{2}{3}P_2(x)

Step 2. Verify by (4): $c_0=\tfrac12\int_{-1}^{1}x^2\,\mathrm{d}x=\tfrac12\cdot\tfrac23=\boxed{\tfrac13}$ ✓ and

c_2=\tfrac{5}{2}\int_{-1}^{1}x^2\cdot\tfrac12(3x^2-1)\,\mathrm{d}x=\tfrac{5}{4}\left(3\cdot\tfrac{2}{5}-\tfrac{2}{3}\right)=\tfrac{5}{4}\cdot\tfrac{8}{15}=\boxed{\tfrac{2}{3}} \ \checkmark

while $c_4=\tfrac92\int x^2P_4\,\mathrm{d}x=0$ — orthogonality refuses to let a degree-4 mode see degree-2 data. Numerical quadrature returns $0.333333$ , $0.666667$ , $0.000000$ .

Step 3. Spot-check at $x=\tfrac12$ : $\tfrac13+\tfrac23P_2(\tfrac12)=\tfrac13-\tfrac1{12}=\tfrac14=(\tfrac12)^2$ ✓. The lesson generalizes: a sphere held at $V\propto\cos^2\theta$ excites only $\ell=0$ and $\ell=2$ — two terms and the problem is over.

2 · Hemispheres at plus and minus V0 (exam level)

A spherical shell of radius $R$ has its northern hemisphere at $+V_0$ , its southern at $-V_0$ . Find the Legendre coefficients of the boundary data and track convergence at $\theta=60°$ .

Step 1. The data is a step in $x=\cos\theta$ : $f=+V_0$ for $x>0$ , $-V_0$ for $x<0$ . It is odd, so only odd $\ell$ survive, and (4) with the parity fold gives $c_\ell=(2\ell+1)V_0\int_0^1P_\ell\,\mathrm{d}x$ .

Step 2. Compute: $\int_0^1x\,\mathrm{d}x=\tfrac12$ gives $c_1=\boxed{\tfrac{3}{2}V_0}$ ; $\int_0^1\tfrac12(5x^3-3x)\,\mathrm{d}x=\tfrac12(\tfrac54-\tfrac32)=-\tfrac18$ gives $c_3=\boxed{-\tfrac{7}{8}V_0}$ ; $\int_0^1\tfrac18(63x^5-70x^3+15x)\,\mathrm{d}x=\tfrac18(\tfrac{63}{6}-\tfrac{70}{4}+\tfrac{15}{2})=\tfrac{1}{16}$ gives $c_5=\boxed{\tfrac{11}{16}V_0}$ . The pattern continues $c_7=-\tfrac{75}{128}V_0$ , $c_9=\tfrac{133}{256}V_0$ : slow, sign-alternating decay (like $\ell^{-3/2}$ ) — the signature of a jump in the data.

Step 3. Evaluate partial sums at $\theta=60°$ , i.e. $x=\tfrac12$ , where the exact surface value is $+V_0$ , using $P_1(\tfrac12)=\tfrac12$ , $P_3(\tfrac12)=-\tfrac{7}{16}$ , $P_5(\tfrac12)=\tfrac{23}{256}$ — surface series in column two; inside at $r=R/2$ (terms damped by $2^{-\ell}$ ) in column three:

through $\ell$	sum at $r=R$ (units of $V_0$ )	sum at $r=R/2$
1	$0.750000$	$0.375000$
3	$1.132812$	$0.422852$
5	$1.194580$	$0.424782$
7	$1.063831$	$0.423760$
9	$0.924650$	$0.423488$

Step 4. Read the table like a physicist. On the surface the sums oscillate stubbornly around the exact value $1$ (through $\ell=41$ the sum is still $1.031$ ) — the Legendre cousin of the Gibbs phenomenon. But inside, $2^{-\ell}$ crushes the high modes: by $\ell=9$ the interior value has converged to $\boxed{V(R/2,60°)=0.4235\,V_0}$ to four decimals. Laplace's equation is a smoothing machine; geometric damping of high- $\ell$ terms is that smoothing made visible.

Origins · 1 minAdrien-Marie Legendre and how this was discovered

Adrien-Marie Legendre

1752–1833 · Paris, the Académie des Sciences

Legendre spent a career being indispensable and under-thanked. Wealthy until the Revolution wiped out his fortune in 1793, he produced first-rank work across an improbable range: the polynomials (born in a 1782 memoir on the attraction of spheroids — precisely this article's problem), quadratic reciprocity, forty years of elliptic integrals, and least squares — published in 1805, then claimed by Gauss as earlier unpublished work, a wound that never closed. Principled to a fault, he refused in 1824 to endorse the government's candidate for the Institut, lost his pension, and died in poverty in 1833. A final indignity: for nearly two centuries, textbooks honoring him printed the face of an unrelated politician. The polynomials, at least, have always carried the right name.

History

1782

— Legendre introduces the polynomials in his memoir on the attraction of spheroids (published 1785).

1785

— Laplace develops the potential theory that will grow into spherical harmonics.

1814

— Gauss constructs his quadrature rule on the roots of Legendre polynomials.

1816

— Olinde Rodrigues, in his doctoral thesis, finds the compact derivative formula now named after him.

1820

— Julien-Léopold Boilly draws his watercolor caricature of Legendre — the only genuine likeness known today.

2009

— The wrong-portrait affair comes to light: for nearly two centuries books had reproduced a portrait of Louis Legendre, an unrelated politician, in place of the mathematician.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.3.2 · Jackson, Classical Electrodynamics, 3rd ed., §3.1–3.3 · Zangwill, Modern Electrodynamics, §7.6–7.7 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 15

Provenance. Equations standard from Griffiths §3.3.2 and Jackson §3.1–3.3. A verifier should: (1) confirm Rodrigues' formula reproduces $P_0..P_4$ as listed; (2) check Bonnet recurrence generates the same polynomials; (3) verify orthogonality integral $2/(2\ell+1)$ (Jackson Eq. 3.21); (4) confirm $P_\ell(1)=1$ , $P_\ell(-1)=(-1)^\ell$ , $P_2(0)=-1/2$ , $P_4(0)=3/8$ ; (5) verify the generating function expansion yields the multipole kernel; (6) check the worked boundary-value result $V_{\rm out}=V_0(R^2/r^2)\cos\theta$ against Griffiths examples. No non-standard numbers used.

Builds on: Laplace's Equation · Separation of Variables in Spherical Coordinates · Boundary Conditions for the Potential Leads to: Associated Legendre Functions · Uniqueness Theorems for the Potential · Image Charges for a Conducting Sphere

Why substitute x equals cos theta at all — why not solve the equation in theta?

You can, but you pay twice. In $\theta$ the equation carries $\sin\theta$ factors in awkward places, the solutions hide their polynomial structure, and the integration weight must be remembered by hand. The substitution maps latitude onto the clean interval $[-1,1]$ , absorbs the weight into $\mathrm{d}x$ , and reveals honest polynomials. When a result is quoted "on $[-1,1]$ ", translate the endpoints back to the poles and $x=0$ to the equator.

What happened to the second solution of the differential equation? Second-order equations have two.

We discarded it deliberately. For each $\ell$ there is a second, non-polynomial solution — the Legendre function of the second kind, $Q_\ell$ — diverging logarithmically at $x=\pm1$ ( $Q_0=\tfrac12\ln\frac{1+x}{1-x}$ is the prototype). On a full sphere the poles are ordinary points of space, so any $Q_\ell$ admixture is inadmissible. In geometries that exclude the polar axis — between coaxial cones, say — the $Q_\ell$ are legal and genuinely needed. Discarding them is a statement about your domain, not the mathematics.

Is a Legendre series just a Fourier series in disguise?

Siblings with the same job — orthogonal decomposition, coefficient-by-integral, completeness — but different habitats. Fourier's sines live on a periodic interval; Legendre polynomials live on an interval whose endpoints are physically special (the poles), where finiteness replaces periodicity as the boundary condition. Consequences: one family suffices, oscillations bunch near the endpoints, the natural quadrature samples non-uniformly. At large $\ell$ , though, $P_\ell(\cos\theta)$ oscillates essentially like a cosine.

How many terms of a Legendre series do I actually need in practice?

A question of smoothness. Polynomial data of degree $n$ needs exactly the modes up to $\ell=n$ — $x^2$ took two terms — and smooth data has coefficients decaying faster than any power of $\ell$ . Discontinuous data is the worst case: the hemisphere coefficients decay like $\ell^{-3/2}$ , and on the sphere the partial sums oscillate for dozens of terms. Off the boundary each term carries $(r/R)^\ell$ or $(R/r)^{\ell+1}$ , so convergence turns geometric — four decimals at $r=R/2$ by $\ell=9$ .

Could a non-integer separation constant sneak through if the series diverges only at the poles?

No. For non-integer effective degree, both independent series solutions diverge logarithmically at $x=\pm1$ , and no combination cancels the blow-up at both endpoints simultaneously. Nor are the poles negotiable: tilt your coordinates and the "bad points" would move, so a legitimate solution must be finite at every candidate pole. The recursion's only escape is exact termination, which happens if and only if the constant is $\ell(\ell+1)$ with integer $\ell\ge0$ — the logic that later quantizes orbital angular momentum.

Where does the square root in the generating function come from? It looks arbitrary.

It is the law of cosines. The distance from a field point at radius $r$ to a source at $r'$ , with angle $\theta$ between them, is $\sqrt{r^2-2rr'\cos\theta+r'^2}$ ; factor out $r$ and you have $r\sqrt{1-2xt+t^2}$ with $x=\cos\theta$ , $t=r'/r$ . The generating function is the Coulomb kernel $1/|\mathbf{r}-\mathbf{r}'|$ made dimensionless — which is why a family defined by an ODE on a sphere also organizes far fields.

Do these polynomials matter outside of electrostatics?

Pervasively. Quantum mechanics: the same equation and finiteness argument appear in every central-potential Schrödinger problem — Legendre quantization is angular momentum quantization. Geophysics: planetary gravity fields are Legendre ladders, Earth's $J_2$ steering every satellite orbit. Numerical analysis: Gauss–Legendre quadrature is the default integrator in finite-element codes. Scattering theory: partial-wave angular distributions are Legendre series in the scattering angle. Learn the family once; four fields greet you by name.