Worked examples — Legendre's equation and Legendre polynomials (intro)

Before we start, one reminder of notation, so nothing is used unearned:

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The scenario matrix

Cell	Case class	What's special	Example
A	Build a new $P_n$ from the recurrence	tests the engine, even parity	Ex 1 ($P_4$)
B	Rodrigues' formula for a specific $n$	alternate route, odd parity	Ex 2 ($P_5$)
C	Degenerate input $n=0$	the "flat pattern", trivial recurrence	Ex 3
D	Endpoint / sign values $x=\pm 1$, parity	boundary + negative argument	Ex 4
E	Orthogonality with $m\ne n$ (different indices)	the integral is zero	Ex 5
F	Orthogonality with $m=n$ (normalization integral)	the integral is nonzero $=\frac{2}{2n+1}$	Ex 6
G	Expand a given polynomial $f(x)$ in the $P_n$ basis	real "Fourier-in-polynomials" use	Ex 7
H	Word problem: potential on a sphere ($x=\cos \theta$)	physics, limiting behaviour at poles	Ex 8
I	Exam twist: non-integer $n$, why no polynomial	degenerate/limiting termination fails	Ex 9

The recurrence engine we lean on repeatedly:

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Cell A — Build $P_4$ from the recurrence

Step 1 — pick the parity. Since $n=4$ is even, use the even series: start from $a_0$, set $a_1=0$. Why this step? The recurrence links coefficients two apart; $n=4$ terminates the even branch at degree 4, so the odd branch would only add junk (and must vanish for a genuine $P_n$).

Step 2 — climb the ladder. With $n=4$, $n(n+1)=20$. Feed $m=0,2,4$ into the engine:

• $a_2=\frac{0\cdot 1-20}{2\cdot 1}{a}_0=\frac{-20}{2}{a}_0=-10a_0.$
• $a_4=\frac{2\cdot 3-20}{4\cdot 3}{a}_2=\frac{-14}{12}{a}_2=-\frac{7}{6}(-10a_0)=\frac{35}{3}{a}_0.$
• $a_6=\frac{4\cdot 5-20}{6\cdot 5}{a}_4=\frac{0}{30}{a}_4=0.$ ✔ terminates.

Why this step? Each rung uses the previous; the numerator hitting $0$ at $m=n=4$ is exactly the "magic termination" — it kills everything of degree $6$ and up.

Step 3 — assemble. $y=a_0(1-10x^2+\frac{35}{3}{x}^4).$

Step 4 — normalize with $P_4(1)=1$. $a_0(1-10+\frac{35}{3})=a_0\cdot \frac{3-30+35}{3}=a_0\cdot \frac{8}{3}=1\Rightarrow a_0=\frac{3}{8}.$ So $P_4(x)=\frac{3}{8}(1-10x^2+\frac{35}{3}{x}^4)=\frac{1}{8}(35x^4-30x^2+3).$

Verify: $P_4(1)=\frac{1}{8}(35-30+3)=\frac{1}{8}\cdot 8=1$ ✔. Matches the parent's forecast answer. Top coefficient positive — check your guess.

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Cell B — Rodrigues' formula for $P_5$ (odd parity)

Step 1 — write the prefactor. $2^5\cdot 5!=32\cdot 120=3840$. Why this step? This constant is what makes $P_n(1)=1$ come out right; getting it wrong scales the whole answer.

Step 2 — expand $(x^2-1{)}^5$ with the binomial theorem. The binomial theorem says $(a+b{)}^k={\sum }_{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})a^{k-j}{b}^j$; here $a=x^2$, $b=-1$, $k=5$, and the coefficients $\left(\genfrac{}{}{0ex}{}{5}{j}\right)$ read off Pascal's row $1,5,10,10,5,1$ with alternating signs from the $(-1{)}^j$: $(x^2-1{)}^5=x^{10}-5x^8+10x^6-10x^4+5x^2-1.$ Why this step? Differentiating a clean polynomial 5 times is far easier than differentiating a product; the binomial theorem gives that clean expansion in one line.

Step 3 — differentiate five times. Term-by-term, $\frac{d^5}{d{x}^5}{x}^k=k(k-1)(k-2)(k-3)(k-4)x^{k-5}$ (and $0$ if $k<5$):

• $x^{10}\to 10\cdot 9\cdot 8\cdot 7\cdot 6x^5=30240x^5$
• $-5x^8\to -5(8\cdot 7\cdot 6\cdot 5\cdot 4)x^3=-5\cdot 6720x^3=-33600x^3$
• $10x^6\to 10(6\cdot 5\cdot 4\cdot 3\cdot 2)x=10\cdot 720x=7200x$
• lower terms vanish.

Why this step? Only terms of degree $\ge 5$ survive five derivatives — this is why the result has degree exactly $n=5$.

Step 4 — divide by $3840$. $P_5(x)=\frac{30240x^5-33600x^3+7200x}{3840}=\frac{1}{8}(63x^5-70x^3+15x).$

Verify: $P_5(1)=\frac{1}{8}(63-70+15)=\frac{1}{8}\cdot 8=1$ ✔. And $P_5(0)=0$ ✔ (odd), matching the forecast.

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Cell C — The degenerate case $n=0$

Step 1 — write the equation with $n=0$. $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }=0.$ Why this step? Plugging $n=0$ removes the undifferentiated term, leaving a first-order equation in $u=y^{\prime }$.

Step 2 — substitute $u=y^{\prime }$. $(1-x^2)u^{\prime }-2xu=0\Rightarrow \frac{u^{\prime }}{u}=\frac{2x}{1-x^2}.$ Why this step? It's now separable; $\frac{2x}{1-x^2}$ is exactly $-\frac{d}{dx}\ln (1-x^2)$, which we can integrate.

Step 3 — integrate. $\ln u=-\ln (1-x^2)+C\Rightarrow u=\frac{K}{1-x^2}.$ Then $y=K\int \frac{dx}{1-x^2}=\frac{K}{2}\ln \frac{1+x}{1-x}+D.$

Step 4 — apply boundedness at the poles. As $x\to \pm 1$, $\ln \frac{1+x}{1-x}\to \pm \infty$. To stay bounded we must set $K=0$, leaving $y=D=$ const. Normalize $P_0(1)=1$: $D=1$. $P_0(x)=1.$

Verify: The discarded piece, up to the constant factor, is $\mathrm{artanh}(x)=\frac{1}{2}\ln \frac{1+x}{1-x}$, whose derivative is $\frac{1}{1-x^2}$ — matching $u$. This is $Q_0(x)$, the log-divergent second solution the parent warned about. Answer to forecast: yes, a second solution exists — we just reject it as unbounded.

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Cell D — Endpoint values and negative argument

Step 1 — use parity at $x=-1$. The rule $P_n(-1)=(-1{)}^n$.

• $P_3(-1)=(-1{)}^3=-1.$
• $P_4(-1)=(-1{)}^4=+1.$

Why this step? $P_n(-1)=P_n(-x){|}_{x=1}=(-1{)}^n{P}_n(1)=(-1{)}^n$. Parity does the work for free — no polynomial evaluation needed.

Step 2 — reflect the whole function. $P_3(-x)=(-1{)}^3{P}_3(x)=-P_3(x)$. Explicitly $-\frac{1}{2}(5x^3-3x)=\frac{1}{2}(3x-5x^3)$ — and indeed $P_3(-x)=\frac{1}{2}(5(-x{)}^3-3(-x))=\frac{1}{2}(-5x^3+3x)$ ✔.

Step 3 — a plain interior value. $P_2(x)=\frac{1}{2}(3x^2-1)$, so $P_2(-\frac{1}{2})=\frac{1}{2}(3\cdot \frac{1}{4}-1)=\frac{1}{2}(\frac{3}{4}-1)=\frac{1}{2}\cdot (-\frac{1}{4})=-\frac{1}{8}.$

Verify: By evenness $P_2(-\frac{1}{2})=P_2(\frac{1}{2})=\frac{1}{2}(\frac{3}{4}-1)=-\frac{1}{8}$ ✔. Forecast: $P_4(-1)=+1$, $P_3(-1)=-1$.

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Cell E — Orthogonality with different indices ($m\ne n$)

Step 1 — invoke orthogonality. ${\int }_{-1}^1{P}_m{P}_{n}dx=\frac{2}{2n+1}{\delta }_{mn}$. Here $m=2\ne n=3$, so ${\delta }_{mn}=0$. Why this step? Different eigenvalues of a Sturm–Liouville problem ⇒ orthogonal eigenfunctions (see Sturm-Liouville Theory). No computation required.

Step 2 — sanity by parity. $P_2$ is even, $P_3$ is odd, so the product is odd; the integral of an odd function over the symmetric interval $[-1,1]$ is $0$. Why this step? It confirms the answer by an independent route — parity, not orthogonality.

${\int }_{-1}^1{P}_2{P}_{3}dx=0.$

Verify: Direct: ${\int }_{-1}^1\frac{1}{2}(3x^2-1)\cdot \frac{1}{2}(5x^3-3x)dx$ integrates an odd polynomial → $0$ ✔.

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Cell F — Orthogonality with equal indices (normalization integral)

Step 1 — square $P_2$. $[\frac{1}{2}(3x^2-1){]}^2=\frac{1}{4}(9x^4-6x^2+1)$. Why this step? We need the explicit integrand to check the theorem, not just quote it.

Step 2 — integrate each power over $[-1,1]$. Using ${\int }_{-1}^1{x}^{2k}dx=\frac{2}{2k+1}$: ${\int }_{-1}^1{x}^4=\frac{2}{5},{\int }_{-1}^1{x}^2=\frac{2}{3},{\int }_{-1}^{1}1=2.$

Step 3 — combine. $\frac{1}{4}(9\cdot \frac{2}{5}-6\cdot \frac{2}{3}+2)=\frac{1}{4}(\frac{18}{5}-4+2)=\frac{1}{4}(\frac{18}{5}-2)=\frac{1}{4}\cdot \frac{8}{5}=\frac{2}{5}.$

Verify: Rule gives $\frac{2}{2(2)+1}=\frac{2}{5}$ ✔. Matches. Also $\frac{2}{5}>0$, as any integral of a square must be.

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Cell G — Expand a given polynomial in the $P_n$ basis

Step 1 — decide which $P_n$ can appear. Degree $\le 2$ and even parity ⇒ only $P_0=1$ and $P_2=\frac{1}{2}(3x^2-1)$. Why this step? The $P_n$ form a basis; a degree-2 even polynomial lives entirely in the span of $\{P_0,P_2\}$, so the sum is finite.

Step 2 — match by algebra (fastest here). Write $x^2=c_0{P}_0+c_2{P}_2=c_0+c_2\frac{1}{2}(3x^2-1)$. Coefficient of $x^2$: $\frac{3}{2}{c}_2=1\Rightarrow c_2=\frac{2}{3}$. Constant: $c_0-\frac{1}{2}{c}_2=0\Rightarrow c_0=\frac{1}{2}{c}_2=\frac{1}{3}$.

Step 3 — cross-check with the projection formula. $c_n=\frac{2n+1}{2}{\int }_{-1}^1{x}^2{P}_{n}dx$. $c_0=\frac{1}{2}{\int }_{-1}^1{x}^{2}dx=\frac{1}{2}\cdot \frac{2}{3}=\frac{1}{3},c_2=\frac{5}{2}{\int }_{-1}^1{x}^2\cdot \frac{1}{2}(3x^2-1)dx.$ The integral $=\frac{1}{2}(3\cdot \frac{2}{5}-\frac{2}{3})=\frac{1}{2}(\frac{6}{5}-\frac{2}{3})=\frac{1}{2}\cdot \frac{8}{15}=\frac{4}{15}$, so $c_2=\frac{5}{2}\cdot \frac{4}{15}=\frac{2}{3}$ ✔. Why this step? Orthogonality turns "find the coefficients" into a single integral each — the Fourier Series idea in polynomial clothing.

$x^2=\frac{1}{3}{P}_0(x)+\frac{2}{3}{P}_2(x).$

Verify: $\frac{1}{3}\cdot 1+\frac{2}{3}\cdot \frac{1}{2}(3x^2-1)=\frac{1}{3}+x^2-\frac{1}{3}=x^2$ ✔.

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Cell H — Word problem: potential outside a hemisphere (physics, $x=\cos \theta$)

How to read the figure. Left panel: a sphere seen edge-on; each coloured spoke is drawn at an angle $\theta$ from the north pole, and its colour encodes the surface potential $V$ there — warm near the top (high $V$) cooling toward the bottom. The two labelled arrows mark the poles where we will read off values. Right panel: the same $V$ plotted against $\theta$; the coral dot at $\theta =0$ sits far higher than the lavender dot at $\theta =\pi$ — a visual proof that the top and bottom are not the same, which is exactly the fingerprint of the odd $P_1$ (dipole) term. If $V$ contained only even modes, the curve would be symmetric about $\theta =\pi /2$; the tilt you see is the dipole.

Step 1 — substitute $x=\cos \theta$. $V=1+2x+\frac{3}{2}{x}^2$. Why this step? In Laplace's Equation in Spherical Coordinates, boundary data naturally comes in powers of $\cos \theta$; converting to $x$ lets us use Legendre machinery directly.

Step 2 — use the Ex 7 expansion for $x^2$. $x^2=\frac{1}{3}{P}_0+\frac{2}{3}{P}_2$, plus $x=P_1$, $1=P_0$: $V=P_0+2P_1+\frac{3}{2}(\frac{1}{3}{P}_0+\frac{2}{3}{P}_2)=(1+\frac{1}{2})P_0+2P_1+P_2.$ $\boxed{V=\frac{3}{2}{P}_0+2P_1+P_2.}$ Why this step? Each $P_n$ is one "multipole mode"; the coefficients are the physical monopole/dipole/quadrupole strengths.

Step 3 — evaluate at the poles using $P_n(\pm 1)$. North pole $x=1$: $V=\frac{3}{2}(1)+2(1)+1=\frac{9}{2}=4.5$. South pole $x=-1$: $V=\frac{3}{2}(1)+2(-1)+1(1)=\frac{3}{2}-2+1=\frac{1}{2}=0.5$.

Verify: Direct from $V=1+2x+\frac{3}{2}{x}^2$: at $x=1$, $1+2+1.5=4.5$ ✔; at $x=-1$, $1-2+1.5=0.5$ ✔. The two poles differ — the $P_1$ (dipole, top-vs-bottom) term breaks the symmetry, exactly what the figure's asymmetric shading shows. Both values are finite — the $P_n$ are bounded at the poles, the physical selection rule at work.

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Cell I — Exam twist: non-integer $n$, why no polynomial

Step 1 — write $n(n+1)$. $n(n+1)=\frac{1}{2}\cdot \frac{3}{2}=\frac{3}{4}$. Why this step? Termination requires some integer $m$ with $m(m+1)=\frac{3}{4}$. But $m(m+1)$ is always an integer for integer $m$ ($0,2,6,12,\dots$), never $\frac{3}{4}$.

Step 2 — compute the first few ratios (even branch $a_0\ne 0$).

• $a_2=\frac{0-\frac{3}{4}}{2}{a}_0=-\frac{3}{8}{a}_0.$
• $a_4=\frac{2\cdot 3-\frac{3}{4}}{12}{a}_2=\frac{21/4}{12}{a}_2=\frac{21}{48}{a}_2=\frac{7}{16}\cdot (-\frac{3}{8})a_0=-\frac{21}{128}{a}_0.$
• $a_6=\frac{4\cdot 5-\frac{3}{4}}{30}{a}_4\ne 0$ — and it never becomes zero.

Why this step? Explicitly seeing $a_6\ne 0$ proves the series is genuinely infinite; there's no clean stopping index.

Step 3 — convergence argument at the endpoints (ratio test). Look at the ratio of successive even coefficients for large $m$: $\frac{a_{m+2}}{a_m}=\frac{m(m+1)-\frac{3}{4}}{(m+2)(m+1)}\underset{m\to \infty }{\to }\frac{m^2}{m^2}=1.$ More precisely the ratio behaves like $\frac{m}{m+2}=1-\frac{2}{m}+\cdots$, so the coefficients shrink only like $1/m$ — far too slowly. Compare with the divergent series $\sum \frac{1}{k}$ at $x=1$: the terms $a_m{x}^m$ at $x=1$ do not go to zero fast enough to be summable, so the series diverges as $x\to \pm 1$. Why this step? This is the reason physics forces $n$ to be a non-negative integer — only then does the numerator vanish, the series stop, and the solution stay finite at the poles.

Verify: Numerically, $a_4/a_0=-\frac{21}{128}\approx -0.1641$, and the ratio $\frac{a_{m+2}}{a_m}=\frac{m(m+1)-3/4}{(m+2)(m+1)}\to 1$ as $m\to \infty$ — the tell-tale of a boundary-divergent series. Non-integer $n$ ⇒ no polynomial ✔.

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Recall Self-test: match example to matrix cell

Which cell is "the answer is zero by parity"? ::: Cell E ($m\ne n$ orthogonality). Which cell shows the second solution $Q_0$ explicitly? ::: Cell C ($n=0$ degenerate case). Which cell proves non-integer $n$ gives no polynomial? ::: Cell I. Which coefficients express $x^2$? ::: $x^2=\frac{1}{3}{P}_0+\frac{2}{3}{P}_2$ (Cell G).

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Connections

• Power Series / Frobenius Method — the recurrence engine used in Cells A, C, I.
• Sturm-Liouville Theory — why the orthogonality of Cells E, F holds.
• Laplace's Equation in Spherical Coordinates — origin of the Cell H potential problem.
• Fourier Series — the "expand in a basis" idea of Cell G.
• Generating Functions — an alternate route to the same $P_n$.
• Associated Legendre Functions & Spherical Harmonics — the next step after this intro.