Frobenius & Special Functions

The lowest-order coefficient gives the indicial equation — for Bessel, $r^2 - \nu^2 = 0$, roots $\pm\nu$ — and each admissible $r$ seeds its own series. The pure case is Cauchy–Euler $x^2y'' + \alpha xy' + \beta y = 0$, where $y = x^r$ alone works and the indicial equation $r(r-1) + \alpha r + \beta = 0$ is the whole story.

Why a decaying cosine: substituting $y = u/\sqrt x$ into Bessel's equation gives $u'' + \left(1 + \dfrac{1/4 - \nu^2}{x^2}\right)u = 0$. For large $x$ the correction term fades, $u$ behaves like $\cos(x - \varphi)$, and so $y$ oscillates with amplitude shrinking like $1/\sqrt x$. The zeros are not equally spaced — gaps $5.520 - 2.405 = 3.115$, then $8.654 - 5.520 = 3.134$ — but the gaps approach $\pi$ as the correction dies out. For $\nu = \tfrac12$ the correction is exactly zero, which is why $J_{1/2}$ is elementary.

Why boundedness truncates: with parameter $\lambda$ in place of $n(n+1)$, the series recurrence is $a_{k+2} = \dfrac{k(k+1) - \lambda}{(k+2)(k+1)}\,a_k$. If the numerator never vanishes, the ratio $a_{k+2}/a_k$ tends to $1$ and the series behaves like $1 + x^2 + x^4 + \cdots$ near $x = \pm1$ — the solution blows up at the poles. The only escape is a numerator that hits zero, $k(k+1) = \lambda$, and then one of the two series stops and becomes a polynomial. For $\lambda = 6$: $a_2 = -3a_0$, $a_4 = 0$, giving $1 - 3x^2$, a multiple of $P_2$.

Why exactly these three: the substitution $x = e^t$ turns $xy'$ into $\dfrac{dy}{dt}$ and $x^2y''$ into $\dfrac{d^2y}{dt^2} - \dfrac{dy}{dt}$, so the equation becomes $\ddot y + (\alpha - 1)\dot y + \beta y = 0$ with constant coefficients, whose characteristic equation $m^2 + (\alpha - 1)m + \beta = 0$ is the indicial equation. The familiar $e^{mt}$, $te^{mt}$ and $e^{at}\cos bt$ become $x^r$, $x^r\ln x$ and $x^a\cos(b\ln x)$ once $t = \ln x$.

Step 1 — two distinct real roots give $y = Ax^2 + Bx^{-2}$.

Step 2 — the data: $y(1) = A + B = 2$, and $y' = 2Ax - 2Bx^{-3}$ gives $y'(1) = 2A - 2B = 0$. So $A = B = 1$ and $y = x^2 + \dfrac1{x^2}$.

Step 3 — check: for $y = x^{-2}$, $x^2\cdot6x^{-4} + x\cdot(-2x^{-3}) - 4x^{-2} = (6 - 2 - 4)x^{-2} = 0$, and $x^2$ works the same way.

Step 4 — read the graph: the $x^{-2}$ part dominates near the singular point $x = 0$ and the $x^2$ part far out; the solution is smallest at $x = 1$, where the initial slope was $0$. This solution cannot be continued through $x = 0$ — the singular point is a genuine wall.

Step 1 — interpret $x^{2i}$: $x^{2i} = e^{2i\ln x} = \cos(2\ln x) + i\sin(2\ln x)$, so the real solutions are $\cos(2\ln x)$ and $\sin(2\ln x)$, and $y = A\cos(2\ln x) + B\sin(2\ln x)$.

Step 2 — the data: $y(1) = A = 1$; $y' = \dfrac{2}{x}\big(-A\sin(2\ln x) + B\cos(2\ln x)\big)$, so $y'(1) = 2B = 0$. Hence $y = \cos(2\ln x)$.

Step 3 — the zeros are where $2\ln x = \tfrac\pi2 + k\pi$, that is $x = e^{\pi/4 + k\pi/2}$: $\ldots, 0.095, 0.456, 2.193, 10.55, \ldots$ Each zero is $e^{\pi/2} \approx 4.81$ times the previous one — equally spaced in $\ln x$, not in $x$.

Step 4 — toward the singular point the zeros pile up without end, so the solution oscillates infinitely often on $(0, 1)$ while staying bounded by $1$.

Why the lowest power decides $r$: near $0$ the equation looks like its Cauchy–Euler part $x^2y'' + p(0)xy' + q(0)y = 0$, whose solutions are $x^r$ with $r$ from the same quadratic; the higher coefficients of $p$ and $q$ only multiply $x^r$ by a power series. Trouble can come only when $r_2 + N = r_1$ for a whole number $N$: at step $N$ the recurrence for the $r_2$ series divides by the indicial polynomial evaluated at $r_2 + N = r_1$, which is zero.

Step 1 — substitute $y = \sum a_nx^{n+r}$ into $4xy'' + 2y' + y$. The derivative terms produce $x^{n+r-1}$ and the $y$ term $x^{n+r}$, so the coefficient of $x^{n+r-1}$ is $\big[4(n+r)(n+r-1) + 2(n+r)\big]a_n + a_{n-1} = 0$, that is $2(n+r)(2n + 2r - 1)\,a_n = -a_{n-1}$.

Step 2 — root $r = 0$: $a_n = -\dfrac{a_{n-1}}{2n(2n-1)}$, so $a_1 = -\tfrac12$, $a_2 = \tfrac1{24}$: $y_1 = 1 - \tfrac x2 + \tfrac{x^2}{24} - \cdots$, which is $\cos\sqrt x$ (put $u = \sqrt x$ in $1 - \tfrac{u^2}{2!} + \tfrac{u^4}{4!}$).

Step 3 — root $r = \tfrac12$: $a_n = -\dfrac{a_{n-1}}{(2n+1)(2n)}$, so $a_1 = -\tfrac16$, $a_2 = \tfrac1{120}$: $y_2 = x^{1/2}\left(1 - \tfrac x6 + \tfrac{x^2}{120} - \cdots\right) = \sin\sqrt x$.

Step 4 — the roots differ by $\tfrac12$, not an integer, so both series exist and $y = c_1\cos\sqrt x + c_2\sin\sqrt x$. The second solution has a vertical tangent at $0$ — something no ordinary power series can have.

Step 1 — take the smaller root $r = -\tfrac12$, the one that might fail. Then $(n - \tfrac12)^2 - \tfrac14 = n(n - 1)$ and the recurrence reads $n(n-1)a_n = -a_{n-2}$.

Step 2 — the danger point is $n = 1$: the equation is $0\cdot a_1 = -a_{-1} = 0$, which holds for every $a_1$. No contradiction, so no logarithm; instead $a_1$ is a second free constant.

Step 3 — from $a_0 = 1$: $a_2 = -\tfrac12$, $a_4 = \tfrac1{24}$, giving $x^{-1/2}\left(1 - \tfrac{x^2}{2} + \tfrac{x^4}{24} - \cdots\right) = \dfrac{\cos x}{\sqrt x}$; from $a_1 = 1$: $a_3 = -\tfrac16$, $a_5 = \tfrac1{120}$, giving $\dfrac{\sin x}{\sqrt x}$.

Step 4 — so $y = \dfrac{A\cos x + B\sin x}{\sqrt x}$, and the bounded solution $\dfrac{\sin x}{\sqrt x}$ is a multiple of $J_{1/2}(x) = \sqrt{\tfrac{2}{\pi x}}\sin x$. Had the right side at $n = 1$ been nonzero, the $r = -\tfrac12$ series would have broken down, and the second solution would have needed a $\ln x$ term.

Since $x^r \ne 0$ for $x > 0$, the bracket must vanish — a quadratic in $r$ whose two roots give two power solutions (a double root contributes $x^r\ln x$, precisely parallel to the $xe^{rx}$ of constant-coefficient theory, via the substitution $x = e^t$ which converts one theory into the other). $\blacksquare$

The $x^{2k+1}$ coefficient is $c_{k+1}(2k+2)(2k+1) + c_{k+1}(2k+2) + c_k = c_{k+1}(2k+2)^2 + c_k$, and indeed $c_{k+1} = \tfrac{-c_k}{4(k+1)^2}$ makes it vanish: $\tfrac{-c_k}{4(k+1)^2}\cdot4(k+1)^2 + c_k = 0$. Every coefficient dies — the series solves the equation exactly. $\blacksquare$

Step 1 — try $y = \sum a_nx^{n+r}$; the lowest-order terms give the indicial equation $2r(r-1) + r = r(2r - 1) = 0$: $r = 0$ or $\tfrac12$.

Step 2 — for $r = \tfrac12$, matching the coefficient of $x^{n+r}$ gives the recurrence $a_{n+1} = -\dfrac{a_n}{(n+1)(2n+3)}$ (from $2(n+r+1)(n+r)a_{n+1} + (n+r+1)a_{n+1} + a_n = 0$ with $r = \tfrac12$).

Step 3 — run it from $a_0 = 1$: $a_1 = -\tfrac{1}{1\cdot3} = -\tfrac13$, $a_2 = \tfrac{1/3}{2\cdot5} = \tfrac{1}{30}$.

Step 4 — the solution: $y = x^{1/2}\left(1 - \tfrac x3 + \tfrac{x^2}{30} - \cdots\right)$ — a square-root prefactor no ordinary power series could produce, and the recurrence generates as many terms as accuracy demands.

Step 1 — indicial equation: $r(r-1) - 3r + 4 = r^2 - 4r + 4 = (r-2)^2$: double root $r = 2$.

Step 2 — double root means $y = x^2(A + B\ln x)$ (the $\ln x$ plays the role $t$ plays for repeated constant-coefficient roots).

Step 3 — data: $y(1) = A = 1$; $y' = 2x(A + B\ln x) + Bx$, so $y'(1) = 2A + B = 0$, $B = -2$.

Step 4 — solution $y = x^2(1 - 2\ln x)$.

Step 5 — verify at $x = e$: $y = e^2(1 - 2) = -e^2$; direct substitution of $y = x^2 - 2x^2\ln x$ into the equation collapses to $0$ term by term ✓.