Power series solutions — ordinary points: Visual walkthrough

We solve the honest little equation $y^{''} + y = 0$ about the point $x_{0} = 0$ , and we watch every symbol earn its place. See the parent Power Series Solutions — Ordinary Points for the full recipe and the mistake list.

Step 1 — What is a power series, and why guess one?

WHAT. A power series is just an infinitely long polynomial: $y (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots = \sum_{n = 0}^{\infty} a_{n} x^{n} .$

Read each symbol where it sits:

$x$ — the input, a slider you can move left/right from $0$ .
$a_{n}$ — a fixed number (unknown for now), the "weight" of the term $x^{n}$ .
$a_{n} x^{n}$ — one bent curve; higher $n$ bends harder and only wakes up when $x$ is not tiny.
$\sum_{n = 0}^{\infty}$ — "add all these bent curves together," starting from $n = 0$ .

WHY guess this. Near $x = 0$ , the powers pile up in size order: $1$ dominates, then $x$ , then $x^{2}$ , ... So the first few numbers $a_{0}, a_{1}, a_{2}$ decide the shape near the origin. If we can find those numbers, we know the curve. A smooth curve = a stack of these simple pieces.

PICTURE. Watch how adding one more term nudges the curve closer to a real function.

Figure — Power series solutions — ordinary points

Step 2 — What differentiation does to the stack

WHAT. To use the equation $y^{''} + y = 0$ we need $y^{''}$ , the second derivative (how the slope itself changes — the "curviness"). Differentiating one term uses the power rule: $\frac{d}{d x} a_{n} x^{n} = n a_{n} x^{n - 1} .$

Term by term:

$n$ (the front factor) — falls down from the exponent and multiplies the coefficient.
$x^{n - 1}$ — the exponent drops by one, so the whole series shifts down one power.

Do it twice: $y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}, y^{''} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} .$

WHY the start index moves. In $y^{'}$ the $n = 0$ term was the constant $a_{0}$ ; its derivative is $0$ , so it drops out — the sum now starts at $n = 1$ . In $y^{''}$ the $n = 0$ and $n = 1$ terms both die, so it starts at $n = 2$ . The lower limit is not decoration; forgetting it corrupts your first coefficients (see the parent's re-indexing mistake).

PICTURE. Each derivative slides the coefficient labels one slot to the left and scales them.

Step 3 — Substitute: turn the ODE into a sum-equals-zero

WHAT. The equation says $y^{''} + y = 0$ . Drop in our two series: $this is y^{''} n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} + this is y n = 0 \sum \infty a_{n} x^{n} = 0.$

Left sum — carries powers $x^{0}, x^{1}, x^{2}, \dots$ but written as $x^{n - 2}$ .
Right sum — carries the same powers but written as $x^{n}$ .
$= 0$ — this must hold for every value of $x$ on the slider, not just one.

WHY this is progress. A differential equation (about slopes and curviness) has become a statement about a single infinite sum being zero. That is an algebra problem now.

PICTURE. The two sums are two rows of boxes; they carry the same powers but with mismatched labels $n$ .

Step 4 — Re-index so both rows count the same power

WHAT. In the first sum the power is $x^{n - 2}$ ; rename the dummy counter with $k = n - 2$ , i.e. $n = k + 2$ . Wherever an $n$ appears, replace it: $n (n - 1) a_{n} x^{n - 2} ⟶ (k + 2) (k + 1) a_{k + 2} x^{k} .$

$n \to k + 2$ in the factor: $n (n - 1) \to (k + 2) (k + 1)$ .
$a_{n} \to a_{k + 2}$ : the coefficient index rides along.
$x^{n - 2} \to x^{k}$ : now it plainly carries $x^{k}$ .
Start moves: $n = 2$ becomes $k = 0$ .

The second sum already carries $x^{n}$ ; just rename $n \to k$ . Both now read $\sum_{k = 0}^{\infty} (\dots) x^{k}$ : $\sum_{k = 0}^{\infty} [(k + 2) (k + 1) a_{k + 2} + a_{k}] x^{k} = 0.$

WHY. You may only add two series coefficient-by-coefficient when they list the same powers. Re-indexing is the alignment that makes that legal.

PICTURE. The re-labelled boxes now sit in matching columns, ready to be added.

Step 5 — The coefficient of every power must be zero

WHAT. A polynomial (finite or infinite) equals zero for all $x$ only if each coefficient is zero. So for every $k$ : $(k + 2) (k + 1) a_{k + 2} + a_{k} = 0.$

Solve for the new coefficient: $a_{k + 2} = - \frac{a _{k}}{( k + 2 ) ( k + 1 )}$

Read it:

$a_{k + 2}$ — the coefficient we are about to discover.
$a_{k}$ — a coefficient two steps behind, already known.
$- \frac{1}{( k + 2 ) ( k + 1 )}$ — the shrink-and-flip factor: negative (from the $+ a_{k}$ ), and denominator grows, so terms get small fast.

This is a recurrence relation — see Recurrence relations.

WHY "two steps behind." The relation jumps $k \to k + 2$ , so evens only talk to evens and odds only to odds. Two separate chains.

PICTURE. A ladder: each rung is built from the rung two below it, split into an even ladder and an odd ladder.

Step 6 — $a_{0}$ and $a_{1}$ are FREE — the two loose ends

WHAT. The recurrence needs an $a_{k}$ two behind. But $a_{0}$ has nothing two behind it, and neither does $a_{1}$ . So the machine never fixes $a_{0}$ or $a_{1}$ — they are chosen by us.

WHY exactly two. A second-order ODE (highest derivative $y^{''}$ ) always needs two constants of integration. Here they appear as the two ladder-bottoms: $a_{0}$ starts the even ladder, $a_{1}$ starts the odd ladder. This is guaranteed by Existence and uniqueness for linear ODEs.

PICTURE. Two ladders standing on two free feet, $a_{0}$ and $a_{1}$ .

Step 7 — Crank the ladders and recognise the answer

WHAT. Feed the recurrence $a_{k + 2} = - \frac{a _{k}}{( k + 2 ) ( k + 1 )}$ .

Even chain (from $a_{0}$ ): $a_{2} = - \frac{a _{0}}{2 \cdot 1} = - \frac{a _{0}}{2 !}, a_{4} = - \frac{a _{2}}{4 \cdot 3} = + \frac{a _{0}}{4 !}, a_{6} = - \frac{a _{0}}{6 !}, \dots$

Odd chain (from $a_{1}$ ): $a_{3} = - \frac{a _{1}}{3 \cdot 2} = - \frac{a _{1}}{3 !}, a_{5} = - \frac{a _{3}}{5 \cdot 4} = + \frac{a _{1}}{5 !}, \dots$

Collect by which foot they stand on: $y = a_{0} = c o s x (1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots) + a_{1} = s i n x (x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots) .$

The alternating signs come from the minus in the recurrence, flipping every step.
The factorials come from the growing denominator $(k + 2) (k + 1)$ multiplying up.

WHY this is a triumph. We never assumed $cos$ or $sin$ . The equation manufactured them, coefficient by coefficient. This connects directly to Taylor series and analyticity — those famous series are exactly what the ODE forces. (For an equation whose series stops early instead, giving a polynomial, see Legendre's equation and Legendre polynomials.)

PICTURE. The two collected columns, revealed as $cos x$ (blue) and $sin x$ (pink).

Step 8 — The degenerate edge: what if the recurrence numerator hits zero?

WHAT. In our equation $y^{''} + y = 0$ the factor is always $- \frac{1}{( k + 2 ) ( k + 1 )}$ , which is never zero — so both ladders run forever (two genuine infinite series). But you must know the other case. In a Legendre-type equation the recurrence looks like $a_{k + 2} = \frac{( k + 3 ) ( k - 2 )}{( k + 2 ) ( k + 1 )} a_{k},$ and at $k = 2$ the numerator $(k - 2) = 0$ , so $a_{4} = 0$ and the whole even ladder past that rung collapses to zero. That branch becomes a finite polynomial.

WHY it matters. Whenever a numerator factor can vanish for some $k$ , that ladder terminates. That is the mechanism behind Legendre and Hermite polynomials — the series is engineered to stop.

PICTURE. Two ladders side by side: ours (endless) vs. a terminating one where a rung snaps.

The one-picture summary

Everything — guess, differentiate, substitute, re-index, kill coefficients, crank the two ladders, harvest $cos$ and $sin$ — compressed into a single flow.

Recall Feynman retelling — say it to a friend with no notation

We wanted a curve $y$ obeying "your curviness is the negative of your height" — that's what $y^{''} + y = 0$ means. We don't know a normal function that does this, so we guess the curve is a giant polynomial with unknown weights $a_{0}, a_{1}, a_{2}, \dots$ Taking the derivative of a polynomial just slides all the weights over and scales them. When we slide $y$ twice and add it back to itself and demand the result be flat-zero everywhere, we get one rule: each weight two ahead equals minus the weight behind, divided by two growing numbers. That rule can't reach back before $a_{0}$ or $a_{1}$ , so those two we pick ourselves — the two dials of a second-order equation. Turning the crank, the even dial prints $1, - \frac{1}{2}, + \frac{1}{24}, \dots$ which is exactly $cos x$ , and the odd dial prints $x, - \frac{1}{6} x^{3}, \dots$ which is $sin x$ . The equation built sine and cosine for us out of pure bookkeeping. And if the little factory rule ever had a "times zero" moment, one ladder would stop early and give a plain polynomial instead — which is how Legendre and Hermite polynomials are born.