4.6.17 · D5Ordinary Differential Equations
Question bank — Power series solutions — ordinary points
Before we start, three words we lean on constantly — pinned so no symbol is unearned:
True or false — justify
A rational function is analytic at exactly when
True — a ratio of polynomials has a convergent Taylor series everywhere its denominator is nonzero; the nearest zero of is the nearest singularity.
If the leading coefficient in is nonzero at , then is an ordinary point
True (for polynomial coefficients) — dividing by keeps and finite and analytic there, so both are analytic.
Every ordinary point yields exactly two linearly independent power-series solutions
True — this is Fuchs' theorem's promise at an ordinary point; and seed the two independent branches (see Existence and uniqueness for linear ODEs).
The radius of convergence of the series solution can be smaller than the distance to the nearest singular point
False — Fuchs guarantees it is at least that distance; it can be larger (if a singularity is removable) but never guaranteed smaller.
A power series solution about an ordinary point always converges for all real
False — only if there is no finite singular point (e.g. Hermite/Airy). For the odd series about converges only for .
If a series is identically zero on an interval, some of its coefficients could still be nonzero
False — a convergent power series that is zero on an interval has every coefficient zero; this is the identity that licenses "set each coefficient to ".
The recurrence relation determines and once you know the ODE
False — the recurrence links to earlier coefficients, so it never touches ; they remain the two free integration constants.
You may add two series term-by-term even if their powers of are offset
False — coefficients can only be combined power-by-power; you must re-index so every sum reads first (see Recurrence relations).
Spot the error
" is singular for because appears."
Error — classification uses , ; at the denominator is , so is ordinary. Singularities are at .
"I substitute straight into , so I never need standard form."
Half-error — you substitute into the undivided form (fine), but you must classify the point via the standard-form first, or you cannot know the point is ordinary or where convergence stops.
"Re-index in gives ."
Error — the lower limit shifts too: , so the sum starts at . Keeping silently drops the and terms and corrupts .
"."
Error — the and terms carry the factor , so they contribute nothing; the sum honestly starts at . Writing isn't wrong in value but invites index-mismatch mistakes.
"The recurrence makes both series terminate because gives zero."
Error — zeroes only the even branch (through ). The odd branch runs on odd , and never hits zero there, so it is an infinite series.
"After substituting I got , so I set the whole bracket-sum to zero."
Error — you set each coefficient bracket to zero individually, one equation per . Summing them into one equation loses all the information needed for the recurrence.
"Since are free, I can also choose freely."
Error — once is chosen, is forced by the recurrence ( for ). Only the first two are free; a 2nd-order ODE has exactly two constants.
Why questions
Why must we re-index at all instead of just adding the sums as written?
Because contribute powers under the same label ; re-indexing renames the dummy so identical powers line up and can legally be added coefficient-by-coefficient.
Why is analyticity of and (not ) the thing we check?
The theorem's hypothesis is on the coefficients of the equation; if are analytic, the conclusion is that a well-behaved analytic solution exists. We cannot check in advance — that's what we're solving for.
Why does substituting a power series turn a differential equation into algebra?
Differentiating only shifts and scales indices, so the ODE becomes a relation between coefficients — the recurrence — with no derivatives left (this is why the whole method exists).
Why do Legendre-type equations sometimes give polynomial (terminating) solutions?
The recurrence numerator contains a factor like ; when the next coefficient becomes zero and the chain stops, leaving a finite polynomial — the Legendre polynomials.
Why can't we use this ordinary-point recipe at a singular point?
At a singular point or is not analytic, so the plain series may not converge or may miss a solution; you need the fractional/log powers of the Singular points and Frobenius method.
Why do Hermite and Airy equations always give solutions convergent for all ?
Their standard-form are polynomials, hence analytic on the entire real line with no finite singular point, so the guaranteed radius of convergence is infinite (see Hermite and Airy equations).
Why keep polynomial coefficients like inside the sums during substitution?
Multiplying a series by shifts its powers; distributing the coefficient first, then re-indexing each resulting sum, keeps every term's power explicit and prevents dropped terms.
Edge cases
Is automatically ordinary for every ODE?
No — only if are analytic there. For , blows up, so is singular despite looking innocent.
What if is analytic but is not?
Still a singular point — the definition requires both and analytic; one failure is enough to disqualify the point.
What happens if one branch terminates and the other doesn't?
You get one polynomial solution and one genuine infinite series; together they are still two linearly independent solutions, exactly as Fuchs promises.
If the nearest singular point is a complex number, does it affect a real series?
Yes — the radius of convergence is the distance to the nearest singularity in the complex plane, so a complex singular point at caps convergence at even on the real line.
What if ? Does the method change?
No — substitute so the ordinary point sits at , expand , run the same recipe, then translate back.
Can a first-order ODE be solved this way, and how many constants appear?
Yes, with a single free constant — the recurrence fixes from , leaving only free, matching a first-order equation's one constant.
What if both and ?
Then the recurrence forces every , giving the trivial solution — consistent, since the only solution with for a linear homogeneous ODE is zero.
Recall One-line summary of the traps
Classify by == in standard form, re-index including the lower limit, set each coefficient to zero, and remember only are free. Convergence reaches the nearest (possibly complex) singularity==.