Exercises — Legendre's equation and Legendre polynomials (intro)

WHAT/WHY: We use Rodrigues' formula, because it expresses $P_n$ explicitly and makes the sign-flip $x\to -x$ trackable. Let $g(x)=(x^2-1{)}^n$. Since $(-x{)}^2=x^2$, we have $g(-x)=g(x)$ — $g$ is an even function. Rodrigues gives $P_n(x)=\frac{1}{2^{n}n!}{g}^{(n)}(x)$, the $n$-th derivative of $g$. Key fact about derivatives: each derivative flips parity. If $g$ is even, $g^{\prime }$ is odd, $g^{\prime \prime }$ is even, and in general $g^{(n)}(-x)=(-1{)}^n{g}^{(n)}(x)$. (Chain rule: differentiating $g(-x)$ brings down one factor of $-1$ per derivative.) Therefore $P_n(-x)=\frac{1}{2^{n}n!}{g}^{(n)}(-x)=(-1{)}^n\frac{1}{2^{n}n!}{g}^{(n)}(x)=(-1{)}^n{P}_n(x).■$ Consequence: even $n$ $\Rightarrow$ $P_n$ even (only even powers); odd $n$ $\Rightarrow$ $P_n$ odd (only odd powers). We now use this freely.

WHAT/WHY: The normalization convention is $P_n(1)=1$ for every $n$, so no computation is needed. The parity fact $P_n(-x)=(-1{)}^n{P}_n(x)$ (Problem 0) tells us $P_n(-1)=(-1{)}^n$.

• $P_5(1)=1$.
• $P_5(-1)=(-1{)}^5=-1$.
• $n=5$ is odd $\Rightarrow$ $P_5$ is an odd function. Sanity check against the explicit form $P_5=\frac{1}{8}(63x^5-70x^3+15x)$: at $x=1$, $\frac{1}{8}(63-70+15)=\frac{1}{8}(8)=1$. ✔ Only odd powers appear, confirming oddness.

WHAT: Write the equation with the $y^{\prime \prime }$ coefficient exposed: $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$. A point is singular where the leading coefficient $(1-x^2)$ vanishes. WHY: $1-x^2=0\Rightarrow x=\pm 1$. At those points the ODE loses its "highest-derivative dominance," so power-series behaviour can break.

• Singular points: $x=+1$ and $x=-1$ (they are regular singular points).
• Because $x=\cos \theta$, these are $\theta =0$ and $\theta =\pi$: the north and south poles of the sphere.

WHY $a_1=0$ (the parity argument): the recurrence $a_{m+2}=(\cdots )a_m$ links coefficients two apart, so it splits the solution into an even chain $a_0\to a_2\to a_4\to \cdots$ and a completely separate odd chain $a_1\to a_3\to a_5\to \cdots$. These two chains never mix. By Problem 0, $P_4$ (even $n$) is an even function, containing only even powers of $x$. Even powers all live in the even chain, so the entire odd chain must be absent — i.e. we set its seed $a_1=0$. That leaves the even chain alone. WHAT/WHY the ratios: the recurrence with $n=4$ has numerator $m(m+1)-20$. March $m=0,2,4$:

• $m=0$: $\frac{0-20}{2\cdot 1}=\frac{-20}{2}=-10$, so $a_2/a_0=-10$.
• $m=2$: $\frac{6-20}{4\cdot 3}=\frac{-14}{12}=-\frac{7}{6}$, so $a_4/a_2=-\frac{7}{6}$.
• $m=4$: $\frac{20-20}{6\cdot 5}=0$, so $a_6/a_4=0$. Termination: at $m=n=4$ the numerator is $0$, killing $a_6$ and every higher even coefficient. The surviving powers are $x^0,x^2,x^4$ $\Rightarrow$ degree 4. ✔ matches $P_4$.

WHAT: With $a_2=-10a_0$ and $a_4=-\frac{7}{6}{a}_2=-\frac{7}{6}(-10a_0)=\frac{70}{6}{a}_0=\frac{35}{3}{a}_0$, $y=a_0(1-10x^2+\frac{35}{3}{x}^4).$ WHY normalize: the ODE fixes the shape but leaves one free constant $a_0$; the convention $P_4(1)=1$ pins it. At $x=1$: $y(1)=a_0(1-10+\frac{35}{3})=a_0\cdot \frac{3-30+35}{3}=a_0\cdot \frac{8}{3}$. Setting this $=1$ gives $a_0=\frac{3}{8}$. $P_4(x)=\frac{3}{8}(1-10x^2+\frac{35}{3}{x}^4)=\frac{1}{8}(35x^4-30x^2+3).✔$

WHAT/WHY Rodrigues: it produces $P_n$ by differentiating a known polynomial $(x^2-1{)}^n$ exactly $n$ times, avoiding the recurrence bookkeeping. With $n=3$: $P_3=\frac{1}{2^3\cdot 3!}\frac{d^3}{d{x}^3}(x^2-1{)}^3=\frac{1}{48}\frac{d^3}{d{x}^3}(x^2-1{)}^3$. Expand $(x^2-1{)}^3=x^6-3x^4+3x^2-1$. Differentiate three times:

• $\frac{d}{dx}:6x^5-12x^3+6x$
• $\frac{d^2}{d{x}^2}:30x^4-36x^2+6$
• $\frac{d^3}{d{x}^3}:120x^3-72x$ So $P_3=\frac{1}{48}(120x^3-72x)=\frac{120}{48}{x}^3-\frac{72}{48}x=\frac{5}{2}{x}^3-\frac{3}{2}x=\frac{1}{2}(5x^3-3x).✔$

WHY parity first: $P_1=x$ is odd, $P_2=\frac{1}{2}(3x^2-1)$ is even. Product = odd $\times$ even = odd function. The integral of any odd function over a symmetric interval $[-1,1]$ is $0$ — the left half cancels the right half (see figure). Direct check: ${\int }_{-1}^{1}x\cdot \frac{1}{2}(3x^2-1)dx=\frac{1}{2}{\int }_{-1}^1(3x^3-x)dx.$ Both $3x^3$ and $x$ are odd $\Rightarrow$ each integrates to $0$ on $[-1,1]$. Hence the integral is $0$. ✔ This is orthogonality made concrete.

WHAT: $P_2^2=\frac{1}{4}(3x^2-1{)}^2=\frac{1}{4}(9x^4-6x^2+1)$. ${\int }_{-1}^{1}9x^{4}dx=9\cdot \frac{2}{5}=\frac{18}{5},{\int }_{-1}^{1}6x^{2}dx=6\cdot \frac{2}{3}=4,{\int }_{-1}^{1}1dx=2.$ So ${\int }_{-1}^1{P}_2^2=\frac{1}{4}(\frac{18}{5}-4+2)=\frac{1}{4}(\frac{18}{5}-2)=\frac{1}{4}\cdot \frac{8}{5}=\frac{2}{5}.$ ✔ Meaning: $\frac{2}{2n+1}$ is the "squared length" of $P_n$ in the $L^2[-1,1]$ inner product. It is the divisor you need to extract the coefficient $c_n$ when you expand a function in Legendre polynomials — the direct analogue of dividing by $\pi$ in a Fourier cosine series. Note this is exactly the $m=n$ (diagonal, ${\delta }_{mn}=1$) case of the orthogonality tool.

WHAT: $y=\frac{1}{2}(3x^2-1)$, so $y^{\prime }=3x$, $y^{\prime \prime }=3$. Plug into $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y$ with $n(n+1)=6$: $(1-x^2)(3)-2x(3x)+6\cdot \frac{1}{2}(3x^2-1).$ $=3-3x^2-6x^2+3(3x^2-1)=3-3x^2-6x^2+9x^2-3.$ Collect: constants $3-3=0$; $x^2$ terms $-3-6+9=0$. Total $=0$. ✔ So $P_2$ is a genuine solution, not just a fit.

(a) Matching. Since $P_2=\frac{1}{2}(3x^2-1)$, solve for $x^2$: $x^2=\frac{1}{3}(2P_2+1)=\frac{2}{3}{P}_2+\frac{1}{3}$. And $P_0=1$, so $x^2=\frac{1}{3}{P}_0+\frac{2}{3}{P}_2\Rightarrow c_0=\frac{1}{3},c_2=\frac{2}{3}.$ (b) Orthogonality formula $c_n=\frac{2n+1}{2}{\int }_{-1}^{1}f{P}_{n}dx$.

• $c_0=\frac{1}{2}{\int }_{-1}^1{x}^2\cdot 1dx=\frac{1}{2}\cdot \frac{2}{3}=\frac{1}{3}.$ ✔
• $c_2=\frac{5}{2}{\int }_{-1}^1{x}^2\cdot \frac{1}{2}(3x^2-1)dx=\frac{5}{4}{\int }_{-1}^1(3x^4-x^2)dx=\frac{5}{4}(3\cdot \frac{2}{5}-\frac{2}{3})=\frac{5}{4}(\frac{6}{5}-\frac{2}{3}).$ $\frac{6}{5}-\frac{2}{3}=\frac{18-10}{15}=\frac{8}{15}$, so $c_2=\frac{5}{4}\cdot \frac{8}{15}=\frac{40}{60}=\frac{2}{3}.$ ✔ Both methods give $c_0=\frac{1}{3},c_2=\frac{2}{3}$ — the "Fourier-in-polynomials" idea in action.

WHAT: $x^3{P}_1=x^3\cdot x=x^4$. We want ${\int }_{-1}^1{x}^{4}dx$, but let's do it the "structural" way to show the machinery. Write $x^4$ in Legendre form. From $P_4=\frac{1}{8}(35x^4-30x^2+3)$ we get $35x^4=8P_4+30x^2-3$. Using $x^2=\frac{1}{3}{P}_0+\frac{2}{3}{P}_2$ (Problem 9): $30x^2=10P_0+20P_2$, and $3=3P_0$. So $35x^4=8P_4+10P_0+20P_2-3P_0=8P_4+20P_2+7P_0.$ $x^4=\frac{8}{35}{P}_4+\frac{4}{7}{P}_2+\frac{1}{5}{P}_0.$ WHY this helps: ${\int }_{-1}^1{x}^{4}dx={\int }_{-1}^1{x}^4{P}_{0}dx$, and by orthogonality only the $P_0$-component survives: ${\int }_{-1}^1{x}^{4}dx=\frac{1}{5}{\int }_{-1}^1{P}_0^{2}dx=\frac{1}{5}\cdot 2=\frac{2}{5}.$ Direct check: ${\int }_{-1}^1{x}^{4}dx=\frac{2}{5}$. ✔ Same answer, no brute force — orthogonality did the projection for us.

WHAT: Evaluate the standard form at the singular point $x=1$. There $(1-x^2)=0$, so the $y^{\prime \prime }$ term drops: $0\cdot y^{\prime \prime }(1)-2(1)y^{\prime }(1)+n(n+1)y(1)=0\Rightarrow -2y^{\prime }(1)+n(n+1)y(1)=0.$ This gives the boundary relation $y^{\prime }(1)=\frac{1}{2}n(n+1)y(1)$. WHY it forces nonzero value: suppose $y(1)=0$. Then $y^{\prime }(1)=0$ too. But a bounded polynomial solution of a 2nd-order linear ODE is determined near the regular singular point by its value and derivative there; both being zero forces $y\equiv 0$ (the trivial solution). Since $P_n$ is a nontrivial degree-$n$ polynomial, $y(1)\ne 0$. Therefore we may divide by $y(1)$ to normalize $P_n(1)=1$, and the choice is consistent. $■$ Numeric sanity ($n=2$): $P_2^{\prime }(1)=3\cdot 1=3$ and $\frac{1}{2}\cdot 2\cdot 3\cdot P_2(1)=3\cdot 1=3$. ✔ boundary relation holds.

WHAT (self-adjoint form): the equation is $\frac{d}{dx}[(1-x^2)y^{\prime }]+n(n+1)y=0$. Call $L[y]=\frac{d}{dx}[(1-x^2)y^{\prime }]$. Let $P_m,P_n$ solve it with eigenvalues ${\lambda }_m=m(m+1),{\lambda }_n=n(n+1)$: $L[P_m]=-{\lambda }_m{P}_m,L[P_n]=-{\lambda }_n{P}_n.$ WHY multiply-and-subtract: to isolate $({\lambda }_n-{\lambda }_m)$ times the overlap integral. Multiply the first by $P_n$, the second by $P_m$, subtract, integrate over $[-1,1]$: ${\int }_{-1}^1(P_{n}L[P_m]-P_{m}L[P_n])dx=({\lambda }_n-{\lambda }_m){\int }_{-1}^1{P}_m{P}_{n}dx.$ The left side is a boundary term (Lagrange's identity): $P_{n}L[P_m]-P_{m}L[P_n]=\frac{d}{dx}[(1-x^2)(P_n{P}_m^{\prime }-P_m{P}_n^{\prime }\left)\right].$ (Expand the derivative to verify: the cross terms $(1-x^2)(P_n{P}_m^{\prime \prime }-P_m{P}_n^{\prime \prime })$ and $-2x(\dots )$ reconstruct $P_{n}L[P_m]-P_{m}L[P_n]$.) Integrating a total derivative: $[(1-x^2)(P_n{P}_m^{\prime }-P_m{P}_n^{\prime }){]}_{-1}^1.$ WHERE it dies: the factor $(1-x^2)$ is zero at both endpoints $x=\pm 1$. So the whole boundary term vanishes — this is why the singular points of the ODE are exactly what make it self-adjoint on $[-1,1]$. Hence $({\lambda }_n-{\lambda }_m){\int }_{-1}^1{P}_m{P}_{n}dx=0.$ For $m\ne n$, ${\lambda }_n\ne {\lambda }_m$, so ${\int }_{-1}^1{P}_m{P}_{n}dx=0$. $■$ See Sturm-Liouville Theory for the general theorem this instance illustrates.