Exercises — Power series solutions — ordinary points

WHAT we do: divide by whatever multiplies $y^{\prime \prime }$, then read off $P$ and $Q$ and ask "does each have a convergent power series (is it analytic) at $x=0$?" A ratio of polynomials is analytic wherever its denominator is non-zero. That is the whole test.

(a) Already standard: $P(x)=x$, $Q(x)=1$. Both are polynomials — analytic everywhere. So $x_0=0$ is an ordinary point. ✔ (This is the Hermite-type shape.)

(b) Divide by $(1-x^2)$: $P=\frac{-2x}{1-x^2},Q=\frac{12}{1-x^2}.$ Denominator $1-x^2=0$ only at $x=\pm 1$. At $x=0$ the denominator is $1\ne 0$, so both are analytic there. $x_0=0$ is an ordinary point; $x=\pm 1$ are singular.

(c) Divide by $x$: $P=\frac{1}{x},Q=1.$ $P$ blows up at $x=0$ (division by zero). So $x_0=0$ is a singular point — the power-series-at-ordinary-point method does not apply; you'd need Singular points and Frobenius method.

Step 1 (assume). $y=\sum _{n=0}^{\infty }{a}_n{x}^n$, so $y^{\prime \prime }=\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}$.

Step 2 (substitute). The term $-xy$ multiplies each $x^n$ by $x$, giving $x^{n+1}$: ${\sum }_{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-{\sum }_{n=0}^{\infty }{a}_n{x}^{n+1}=0.$

Step 3 (re-index to a common power $x^k$). Why? We can only add coefficients of the same power. In the first sum let $k=n-2$ (so $n=k+2$, and $n=2\Rightarrow k=0$). In the second let $k=n+1$ (so $n=k-1$, and $n=0\Rightarrow k=1$): ${\sum }_{k=0}^{\infty }(k+2)(k+1)a_{k+2}{x}^k-{\sum }_{k=1}^{\infty }{a}_{k-1}{x}^k=0.$

Step 4 (handle the odd-man-out $k=0$). The second sum starts at $k=1$, so $k=0$ appears only in the first: $k=0:(2)(1)a_2=0\Rightarrow \boxed{a_2=0}.$ For $k\ge 1$, set the coefficient of $x^k$ to zero: $(k+2)(k+1)a_{k+2}-a_{k-1}=0\Rightarrow \boxed{a_{k+2}=\frac{a_{k-1}}{(k+2)(k+1)}},k\ge 1.$

Step 5 (crank).

• $k=1:a_3=\frac{a_0}{3\cdot 2}=\frac{a_0}{6}$
• $k=2:a_4=\frac{a_1}{4\cdot 3}=\frac{a_1}{12}$
• $k=3:a_5=\frac{a_2}{5\cdot 4}=0(\text{since }{a}_2=0)$

So $a_2=0,a_3=\frac{a_0}{6},a_4=\frac{a_1}{12},a_5=0$. The pattern that follows: every third coefficient starting at $a_2$ vanishes ($a_2,a_5,a_8,\cdots =0$), and the solution splits into an $a_0$-branch and an $a_1$-branch, exactly as Fuchs promises.

Substitute $y=\sum a_n{x}^n$. Here $y^{\prime }=\sum n{a}_n{x}^{n-1}$, so $-2x{y}^{\prime }=-2\sum n{a}_n{x}^n$ (the $x$ cancels the $x^{n-1}$ back up to $x^n$): ${\sum }_{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-2{\sum }_{n=1}^{\infty }n{a}_n{x}^n+4{\sum }_{n=0}^{\infty }{a}_n{x}^n=0.$

Re-index only the first sum ($k=n-2$); the other two already carry $x^k$ (rename $n\to k$, and note $-2n{a}_n$ at $n=0$ is $0$ so we can start all at $k=0$): ${\sum }_{k=0}^{\infty }[(k+2)(k+1)a_{k+2}-2k{a}_k+4a_k]x^k=0.$

Combine the $a_k$ terms (they share $x^k$): $-2k+4=-2(k-2)$. Set coefficient to zero: $(k+2)(k+1)a_{k+2}=2(k-2)a_k\Rightarrow \boxed{a_{k+2}=\frac{2(k-2)}{(k+2)(k+1)}{a}_k}.$

Key observation: the numerator $2(k-2)$ is zero at $k=2$. So:

• $a_2=\frac{2(0-2)}{2\cdot 1}{a}_0=\frac{-4}{2}{a}_0=-2a_0$.
• $a_4=\frac{2(2-2)}{4\cdot 3}{a}_2=0$, and every higher even term is $0$.

The even branch terminates: $y_{\text{even}}=a_0\left(1-2x^2\right).$ This is (up to scaling) the Hermite polynomial $H_2(x)=4x^2-2$: indeed $1-2x^2=-\frac{1}{2}{H}_2(x)$. See Hermite and Airy equations.

Odd branch does not terminate (numerator $2(k-2)$ never hits $0$ for odd $k\ge 1$):

• $k=1:a_3=\frac{2(-1)}{3\cdot 2}{a}_1=-\frac{1}{3}{a}_1$
• $k=3:a_5=\frac{2(1)}{5\cdot 4}{a}_3=\frac{2}{20}{a}_3=\frac{1}{10}\left(-\frac{1}{3}\right)a_1=-\frac{1}{30}{a}_1$

$y=a_0(1-2x^2)+a_1\left(x-\frac{1}{3}{x}^3-\frac{1}{30}{x}^5-\cdots \right).$

Why the initial data pins the constants: in a power series $y=a_0+a_{1}x+a_2{x}^2+\cdots$, plugging $x=0$ gives $y(0)=a_0$, and differentiating then setting $x=0$ gives $y^{\prime }(0)=a_1$. So the two free constants are the two pieces of initial data: $a_0=y(0)=1,a_1=y^{\prime }(0)=0.$

Recurrence. With $-2x$ replaced by $+x$: $x{y}^{\prime }=\sum n{a}_n{x}^n$, so ${\sum }_{n=2}^{\infty }n(n-1)a_n{x}^{n-2}+{\sum }_{n=1}^{\infty }n{a}_n{x}^n+{\sum }_{n=0}^{\infty }{a}_n{x}^n=0.$ Re-index the first ($k=n-2$), rename the rest ($n\to k$, the $n=1,0$ starts are absorbed since $k{a}_k+a_k$ at $k=0$ is just $a_0$): ${\sum }_{k=0}^{\infty }[(k+2)(k+1)a_{k+2}+k{a}_k+a_k]x^k=0.$ Combine: $k{a}_k+a_k=(k+1)a_k$, so $(k+2)(k+1)a_{k+2}=-(k+1)a_k\Rightarrow \boxed{a_{k+2}=-\frac{a_k}{k+2}}.$ (One factor $(k+1)$ cancels — a small gift.)

Crank with $a_0=1,a_1=0$. Since $a_1=0$, every odd coefficient is $0$ (each odd term traces back to $a_1$).

• $a_2=-\frac{a_0}{2}=-\frac{1}{2}$
• $a_4=-\frac{a_2}{4}=-\frac{1}{4}\left(-\frac{1}{2}\right)=\frac{1}{8}$
• $a_6=-\frac{a_4}{6}=-\frac{1}{6}\cdot \frac{1}{8}=-\frac{1}{48}$

First four non-zero terms: $y=1-\frac{x^2}{2}+\frac{x^4}{8}-\frac{x^6}{48}+\cdots$

Bonus recognition: the pattern $a_{2m}=\frac{(-1{)}^m}{2^{m}m!}$ gives $y={\sum }_{m=0}^{\infty }\frac{(-1{)}^m}{2^{m}m!}{x}^{2m}={\sum }_{m=0}^{\infty }\frac{(-x^2/2{)}^m}{m!}=e^{-x^2/2}.$ Check: $\frac{1}{2^{1}1!}=\frac{1}{2}$ ✔, $\frac{1}{2^{2}2!}=\frac{1}{8}$ ✔, $\frac{1}{2^{3}3!}=\frac{1}{48}$ ✔. A closed form fell out — a nice reward.

(a) Ordinary point + radius. Standard form: divide by $x^2+1$, $P=\frac{x}{x^2+1},Q=\frac{-1}{x^2+1}.$ The denominator $x^2+1=0$ only at $x=\pm i$ (complex). At $x=0$ it equals $1\ne 0$, so both $P,Q$ are analytic — $x_0=0$ is an ordinary point.

By Fuchs' theorem, the radius of convergence is at least the distance from $x_0=0$ to the nearest singularity. The singularities sit at $x=\pm i$, each a distance $|\pm i-0|=1$ away. So $\boxed{R\ge 1.}$ Why the imaginary singularities count: analyticity and radius of convergence live in the complex plane; a real series still "feels" the nearest complex singularity. See Taylor series and analyticity.

(b) Recurrence. Substitute $y=\sum a_n{x}^n$. The pieces: $\underset{x^2{y}^{\prime \prime }}{\underbrace{\sum n(n-1)a_n{x}^n}}+\underset{y^{\prime \prime }}{\underbrace{\sum n(n-1)a_n{x}^{n-2}}}+\underset{x{y}^{\prime }}{\underbrace{\sum n{a}_n{x}^n}}-\underset{y}{\underbrace{\sum a_n{x}^n}}=0.$ Only the $y^{\prime \prime }$ (bare) sum needs re-indexing ($k=n-2$); the rest carry $x^k$ already: ${\sum }_k[(k+2)(k+1)a_{k+2}+k(k-1)a_k+k{a}_k-a_k]x^k=0.$ Combine the $a_k$ terms: $k(k-1)+k-1=k^2-k+k-1=k^2-1=(k-1)(k+1)$. So $(k+2)(k+1)a_{k+2}=-(k-1)(k+1)a_k\Rightarrow \boxed{a_{k+2}=-\frac{(k-1)}{(k+2)}{a}_k}.$ (The common factor $(k+1)$ cancels.)

(c) The polynomial solution. The numerator $-(k-1)$ is zero at $k=1$. Starting the odd branch from $a_1$: $a_3=-\frac{(1-1)}{3}{a}_1=0,$ so all higher odd terms vanish. The entire odd branch is just $y_1=a_{1}x.$ Taking $a_1=1$ gives the exact solution $\boxed{y_1=x}$. (Quick sanity check: $y_1=x\Rightarrow y_1^{\prime }=1,y_1^{\prime \prime }=0$, and $(x^2+1)(0)+x(1)-x=0$ ✔.)

Even branch with $a_0=1$:

• $k=0:a_2=-\frac{(0-1)}{2}{a}_0=-\frac{-1}{2}=\frac{1}{2}$
• $k=2:a_4=-\frac{(2-1)}{4}{a}_2=-\frac{1}{4}\cdot \frac{1}{2}=-\frac{1}{8}$
• $k=4:a_6=-\frac{(4-1)}{6}{a}_4=-\frac{3}{6}\left(-\frac{1}{8}\right)=\frac{1}{16}$

So the second solution begins $y_2=1+\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^6}{16}-\cdots (a_2,a_4,a_6)=\left(\frac{1}{2},-\frac{1}{8},\frac{1}{16}\right).$ (In fact $y_2=\sqrt{1+x^2}$ — its binomial series matches term for term, and its radius of convergence is exactly $1$, meeting the Fuchs bound with equality.)

Ordinary point test uses which coefficients? ::: $P,Q$ from the standard form $y^{\prime \prime }+P{y}^{\prime }+Qy=0$ — both must be analytic at $x_0$. How do initial conditions fix $a_0,a_1$? ::: $a_0=y(x_0)$, $a_1=y^{\prime }(x_0)$. What caps the radius of convergence? ::: The distance to the nearest singular point, measured in the complex plane (Fuchs). When does a series branch terminate into a polynomial? ::: When the recurrence's numerator hits zero at some index, killing that coefficient and all that follow it in the branch.