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Power series solutions

Substituting an unknown series into a linear equation with variable coefficients, re-indexing so the sums combine, matching coefficients to get a recurrence, and reading the guaranteed radius off the equation.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to decide whether a point is ordinary or singular for a linear second-order equation, substitute a power series and shift indices so that every sum runs over the same power, match coefficients to obtain a recurrence, generate the first coefficients from the initial conditions, and state the interval on which the theorem guarantees convergence.

2. What you already have

You can solve a linear equation with constant coefficients from its characteristic roots, and you have met Taylor series and their radii of convergence. The methods of unit 2 all assumed constant coefficients. Once the coefficients depend on $x$, the characteristic equation has nothing to be the characteristic equation of, and this is what replaces it.

3. Words this lesson uses

4. Solving by unknown coefficients

Most linear equations with variable coefficients have no solution expressible in elementary functions. Airy's equation $y'' - xy = 0$ is about as simple as such an equation gets, and no combination of exponentials, powers and trigonometric functions solves it. The response is to widen what counts as an answer: a power series is a function, and if we can determine its coefficients we have determined the solution.

So write

$$y = \sum_{n \ge 0} a_n x^{n}, \qquad y' = \sum_{n \ge 1} n a_n x^{n-1}, \qquad y'' = \sum_{n \ge 2} n(n-1)a_n x^{n-2},$$

and substitute. The awkward step is that the three sums run over different powers, so shift the last one — replacing $n$ by $n + 2$ — to get

$$y'' = \sum_{n \ge 0} (n+2)(n+1)a_{n+2}x^{n}.$$

Now every sum runs over $x^{n}$ and they can be added into one. A power series is identically zero only when every coefficient is zero, so the single equation becomes one equation per power — a recurrence giving each coefficient from earlier ones.

The theorem. If $x_0$ is an ordinary point, the recurrence has a solution for every choice of $a_0$ and $a_1$, and the resulting series converges at least on $|x - x_0| < R$, where $R$ is the distance from $x_0$ to the nearest singular point. Two free coefficients, for a second-order equation: the two arbitrary constants, arriving in a new costume.

5. Why the point has to be checked first

Consider $x^{2}y'' - 2y = 0$ at $x = 0$. Dividing gives $y'' - (2/x^{2})y = 0$, and $2/x^{2}$ has no power series about the origin — the point is singular.

Substituting an ordinary power series anyway does not fail loudly. It produces a recurrence, the recurrence produces coefficients, and the coefficients produce a series that solves nothing. In this case the actual solutions are $x^{2}$ and $x^{-1}$, and the second has no power series about $0$ at all, so the method could not have found it however carefully it was applied.

That is what makes the check the first step rather than a formality. A method that returns a confident wrong answer is more dangerous than one that refuses, and the only defence is to test the hypothesis before starting. Singular points are not beyond reach — the next lesson gets at the well-behaved ones — but they need a different substitution.

6. Where this goes wrong

Adding sums that run over different powers. Until every sum is over $x^{n}$, collecting coefficients matches $x^{2}$ against $x^{4}$ and produces a recurrence that is simply not the equation's.

Losing the first terms when shifting. The $y'$ sum starts at $n = 1$ and the $y''$ sum at $n = 2$; after shifting they start elsewhere, and the leftover lowest-power terms have to be written out separately rather than swept in.

Dividing by the wrong factor. The recurrence divides by $(n+2)(n+1)$, the product of the two indices above $n$. Neither $n(n-1)$ nor $n^{2}$ is it.

Treating the radius as the true one. The theorem gives a lower bound. The series may converge further; it is guaranteed only to the nearest singularity.

7. The method, step by step, and how to check it

A power series solution about an ordinary point is found by the same routine every time, and each move has a reason.

1. Check the point is ordinary. Divide by the coefficient of $y''$; if $p$ and $q$ have power series about $x_0$, the point is ordinary. Note the distance to the nearest singular point, including complex ones: that is the guaranteed radius.
2. Write the three sums for $y$, $y'$ and $y''$.
3. Bring every sum to $x^{n}$. A factor $x^{k}$ in front of a sum raises its powers by $k$; differentiating lowers them. Shift each index until the power is $x^{n}$, and write down where each sum now starts.
4. Match the lowest powers separately if the sums start at different $n$.
5. Set the general coefficient to zero and solve for the highest-index coefficient. Factor before dividing: factors often cancel, and a factor that can vanish tells you the series may stop.
6. Generate the coefficients from $a_0 = y(x_0)$ and $a_1 = y'(x_0)$.

Why every coefficient must vanish. A power series that is zero for every $x$ near $x_0$ has every coefficient zero: put $x = x_0$ to get $a_0 = 0$, differentiate and put $x = x_0$ again to get $a_1 = 0$, and so on. That is what turns one equation between functions into infinitely many equations between numbers.

Why two coefficients are free. The recurrence determines $a_{n + 2}$ from lower coefficients, so $a_0$ and $a_1$ are never determined. They are the two arbitrary constants of a second-order equation, and the initial conditions fix them.

How to check the answer. Substitute the first few terms back into the equation: the low powers of $x$ must cancel exactly. Check the recurrence by putting in a known case, such as $n = 0$, and comparing with a direct calculation. When the equation has an elementary solution, as $y'' + y = 0$ does, the series should reproduce its Taylor series; and when the recurrence has a factor that can vanish, look for a polynomial solution and check it exactly.

8. In the world: the edge of a rainbow

Near the bright edge of a rainbow, and in several other places in physics, the intensity of light is governed by Airy's equation $y'' = xy$, the equation this lesson solved by series. It has no solution in elementary functions, so its values are computed from exactly the series found here. With $y(0) = 1$ and $y'(0) = 0$, the recurrence $a_{n + 2} = \frac{a_{n - 1}}{(n + 2)(n + 1)}$ gives

$$y = 1 + \frac{x^{3}}{6} + \frac{x^{6}}{180} + \frac{x^{9}}{12\,960} + \cdots.$$

At $x = 0.6$ the terms are $1$, $0.036$, $0.000259$ and $0.0000008$: each is hundreds of times smaller than the one before, so three terms give $y(0.6)$ to six decimal places, $1.036260$. That rapid shrinking is how the lesson's radius of convergence shows up in practice: every point is ordinary, the radius is infinite, and near $0$ a handful of terms is enough.

Further out the terms shrink more slowly and more are needed, and far from $0$ physicists switch to other formulas, which is typical: a series is the tool of choice near its centre, and it is how the tables and the functions built into scientific software were first computed. The same Airy function describes a quantum particle in a uniform field, such as an electron near the surface of a semiconductor, and the dark and bright bands of light near a caustic.

9. In the world: the shape of the Earth's gravity

Around a planet that is not perfectly round, the gravitational potential depends on the angle from the pole as well as the distance, and it is written as a sum of Legendre polynomials, the terminating series solutions of $(1 - x^{2})y'' - 2xy' + n(n + 1)y = 0$ with $x$ the cosine of the angle. For $n = 0, 1, 2$ they are $1$, $x$ and $P_2(x) = \frac{3x^{2} - 1}{2}$, and the last is the one that matters most for the Earth: it describes the bulge at the equator. At the pole, $x = 1$, $P_2 = 1$; at the equator, $x = 0$, $P_2 = -\frac{1}{2}$, so the correction to gravity has opposite signs at the two places.

Satellite orbits are computed with this series, and the $P_2$ term, weighted by a number measured from satellites' motion, is large enough that ignoring it would put a GPS satellite kilometres out of place within days. That the series stops, giving a polynomial, is the result of the recurrence's numerator vanishing, the case this lesson's last example worked through: without it the solution would blow up at the poles, $x = \pm 1$, which are the equation's singular points.

10. In the world: why solvers still use series

Modern software solves most differential equations by stepping numerically, but series remain inside the machinery: special functions such as Airy and Bessel functions are evaluated near the origin from their series, the first step of some high-accuracy solvers is a Taylor series about the starting point, and an engineer checking a numerical answer near an initial point often compares it against the first three terms of the series, which can be written down by hand in a minute from the recurrence.

11. A series solution is an answer, not an approximation of one

The word series suggests something provisional — a few terms, accurate near the point, to be replaced later by the real formula. For an equation like Airy's there is no real formula to be replaced by, and the series is not approximating anything: within its radius of convergence it is the function, exactly, in the same way that the exponential series is the exponential. What is provisional is any partial sum, and the two are worth keeping apart in your head. The related error is to treat the guaranteed radius as a statement about the solution's own behaviour. It is a statement about the equation's singular points, computed before any coefficient exists, and a solution can be perfectly well behaved a long way past where its series about one point stops converging.

12. A series for an equation we can already solve

1. Solve $y'' + y = 0$ by a series about $x = 0$. Write $y$ and the shifted $y''$ as sums over $x^{n}$.

   $y = \displaystyle\sum_{n \ge 0} a_nx^{n}, \qquad y'' = \sum_{n \ge 0}(n + 2)(n + 1)a_{n + 2}x^{n}$

   Every point is ordinary, since the coefficient of $y''$ is $1$.

2. Substitute and combine the two sums into one.

   $\displaystyle\sum_{n \ge 0}\left[(n + 2)(n + 1)a_{n + 2} + a_n\right]x^{n} = 0$

   Both sums run over the same powers, so they add term by term.

3. Set each coefficient to zero and solve for $a_{n + 2}$.

   $a_{n + 2} = -\dfrac{a_n}{(n + 2)(n + 1)}$

   A power series is zero only when every coefficient is. The recurrence skips two at a time.

4. Generate the coefficients from $a_0 = 1$, $a_1 = 0$.

   $a_2 = -\dfrac{1}{2 \cdot 1} = -\dfrac{1}{2}, \quad a_4 = -\dfrac{a_2}{4 \cdot 3} = \dfrac{1}{24}, \quad a_1 = a_3 = \dots = 0$

   Every odd coefficient inherits the zero of $a_1$.

5. Recognise the series.

   $y = 1 - \dfrac{x^{2}}{2!} + \dfrac{x^{4}}{4!} - \cdots = \cos x$

   A known answer confirms the machinery; the nearest singular point is infinitely far away, so it converges everywhere.

13. Airy's equation, which has no elementary answer

1. Solve $y'' - xy = 0$ about $x = 0$. Write $xy$ as a sum over $x^{n}$.

   $xy = \displaystyle\sum_{n \ge 0} a_nx^{n + 1} = \sum_{n \ge 1} a_{n - 1}x^{n}$

   The factor $x$ raises every power, so the index shifts the other way: replace $n$ by $n - 1$.

2. Substitute, with $y''$ shifted as usual.

   $\displaystyle\sum_{n \ge 0}(n + 2)(n + 1)a_{n + 2}x^{n} - \sum_{n \ge 1} a_{n - 1}x^{n} = 0$

   Both sums are now over $x^{n}$, but they start at different $n$.

3. Match $x^{0}$ on its own.

   $2 \cdot 1 \cdot a_2 = 0 \quad\Rightarrow\quad a_2 = 0$

   The second sum has no $x^{0}$ term, so only the first contributes.

4. Match $x^{n}$ for $n \ge 1$ and solve for the highest coefficient.

   $a_{n + 2} = \dfrac{a_{n - 1}}{(n + 2)(n + 1)}$

   The recurrence steps three at a time.

5. Generate the first branch from $a_0 = 1$, $a_1 = 0$.

   $a_3 = \dfrac{a_0}{3 \cdot 2} = \dfrac{1}{6}, \quad a_6 = \dfrac{a_3}{6 \cdot 5} = \dfrac{1}{180}: \qquad y_1 = 1 + \dfrac{x^{3}}{6} + \dfrac{x^{6}}{180} + \cdots$

   Only multiples of three survive, because $a_1 = a_2 = 0$.

6. Generate the second branch from $a_0 = 0$, $a_1 = 1$.

   $a_4 = \dfrac{a_1}{4 \cdot 3} = \dfrac{1}{12}, \quad a_7 = \dfrac{a_4}{7 \cdot 6} = \dfrac{1}{504}: \qquad y_2 = x + \dfrac{x^{4}}{12} + \dfrac{x^{7}}{504} + \cdots$

   Two free coefficients give two independent solutions. Every point is ordinary, so both converge for all $x$: the Airy functions are the answer, not a step towards one.

14. A finite radius, and a series that stops

1. Solve $(1 - x^{2})y'' - 2xy' + 2y = 0$ about $x = 0$. Find the singular points and the guaranteed radius.

   $1 - x^{2} = 0 \;\Rightarrow\; x = \pm 1, \qquad R \ge 1$

   $x = 0$ is ordinary, and the nearest singular point is at distance $1$.

2. Write each term as a sum over $x^{n}$.

   $x^{2}y'' = \displaystyle\sum n(n - 1)a_nx^{n}, \qquad xy' = \sum na_nx^{n}, \qquad y'' = \sum(n + 2)(n + 1)a_{n + 2}x^{n}$

   Multiplying by $x^{2}$ undoes the drop of two powers from differentiating twice.

3. Collect the coefficient of $x^{n}$.

   $(n + 2)(n + 1)a_{n + 2} - n(n - 1)a_n - 2na_n + 2a_n = 0$

   Expand $(1 - x^{2})y'' = y'' - x^{2}y''$ and take each sum's coefficient.

4. Factor the $a_n$ terms and solve for $a_{n + 2}$.

   $n^{2} + n - 2 = (n + 2)(n - 1) \quad\Rightarrow\quad a_{n + 2} = \dfrac{(n + 2)(n - 1)}{(n + 2)(n + 1)}a_n = \dfrac{n - 1}{n + 1}a_n$

   $-n(n - 1) - 2n + 2 = -(n^{2} + n - 2)$, and $n + 2$ cancels.

5. Take the odd branch, $a_0 = 0$, $a_1 = 1$.

   $a_3 = \dfrac{1 - 1}{1 + 1}a_1 = 0 \quad\Rightarrow\quad a_5 = a_7 = \dots = 0, \qquad y_2 = x$

   The factor $n - 1$ vanishes at $n = 1$, so the series stops: a polynomial solution.

6. Check the polynomial solution in the equation.

   $y = x: \quad (1 - x^{2}) \cdot 0 - 2x \cdot 1 + 2x = 0$

   It holds for every $x$, not just inside the radius.

7. Take the even branch, $a_0 = 1$, $a_1 = 0$.

   $a_2 = \dfrac{-1}{1} = -1, \quad a_4 = \dfrac{1}{3}a_2 = -\dfrac{1}{3}, \quad a_6 = \dfrac{3}{5}a_4 = -\dfrac{1}{5}: \qquad y_1 = 1 - x^{2} - \dfrac{x^{4}}{3} - \dfrac{x^{6}}{5} - \cdots$

   Put $n = 0, 2, 4$ in turn. This branch never stops.

8. Read the radius from the coefficients.

   $\dfrac{a_{n + 2}}{a_n} = \dfrac{n - 1}{n + 1} \to 1 \quad\Rightarrow\quad R = 1$

   The ratio test on the powers $x^{2}$ gives convergence for $x^{2} < 1$: exactly the radius the singular points predicted.

15. Your turn: find the recurrence for $y'' - 2xy' - 2y = 0$ about $x = 0$

1. Substitute the series; note the middle sum needs no shift.

   $\displaystyle\sum (n + 2)(n + 1)a_{n + 2}x^{n} - 2\sum na_nx^{n} - 2\sum a_nx^{n} = 0$

   The factor of $x$ cancels the shift that differentiating made.

2. Your turn: work this step out. Its working is at the end of the packet.

   Collect the coefficient of $x^{n}$.

3. Your turn: work this step out. Its working is at the end of the packet.

   Solve for $a_{n + 2}$ and simplify.

16. Guided practice

Put the steps of finding a power series solution of $y'' - 4y = 0$ about $x = 0$ into the order they must be done.

Number the steps in order (write the number in the box):

• Check that the point is ordinary
• Substitute a series and differentiate it twice
• Shift the indices so every sum runs over the same power
• Set the coefficient of each power to zero
• Read off the recurrence and generate the coefficients

17. Guided practice

Complete the worked solution: the power series for $y'' - 10y = 0$ with $y(0) = 5$, $y'(0) = 0$.

1. Substitute a series for $y$ and differentiate twice.

   $y = \displaystyle\sum_{n \ge 0} a_nx^{n}, \qquad y'' = \sum_{n \ge 2} n(n - 1)a_nx^{n - 2}$

   Each derivative brings down the exponent and lowers it by one.

2. Shift the index of $y''$ so it runs over $x^{n}$.

   $y'' = \displaystyle\sum_{n \ge 0} (n + 2)(n + 1)a_{n + 2}x^{n}$

   Replace $n$ by $n + 2$ so both sums have the same power.

3. Substitute, set each coefficient to zero, and solve for $a_{n + 2}$.

   $(n + 2)(n + 1)a_{n + 2} - 10a_n = 0 \quad\Rightarrow\quad a_{n + 2} = \dfrac{10a_n}{(n + 2)(n + 1)}$

   A series is zero only when every coefficient is.

4. Read the free coefficients from the initial conditions.

   $a_0 = y(0) = 5, \qquad a_1 = y'(0) =$ _r

   At $x = 0$ the series equals $a_0$ and its derivative equals $a_1$.

5. Put $n = 0$ into the recurrence.

   $a_2 = \dfrac{10 \cdot a_0}{(0 + 2)(0 + 1)} = \dfrac{10 \cdot 5}{2 \cdot 1} =$ _p

   Divide by the factor $2 \cdot 1$ that re-indexing put there.

6. Put $n = 1$ into the recurrence.

   $a_3 = \dfrac{10 \cdot a_1}{(1 + 2)(1 + 1)} =$ _q

   Every odd coefficient inherits the zero of $a_1$, so only even powers appear.

18. Guided practice

Is $x = 0$ an ordinary point of $x^{2}y'' + y' + y = 0$?

• Yes, it is an ordinary point
• No, it is a singular point

19. Practice

Differentiating $\sum a_n x^{n}$ twice and shifting the index leaves $\sum (n+2)(n+1)a_{n+2}x^{n}$. Fill in the value of the factor $(n+2)(n+1)$ for each $n$.

20. Practice

For $y'' - 4y = 0$ with $y(0) = 2$ and $y'(0) = 0$, what is the coefficient $a_2$ of the power series solution about $x = 0$?

Answer:

21. Practice

Find the power series solution of $y'' - 2xy' - 2y = 0$ about $x = 0$ with $y(0) = 16$ and $y'(0) = 0$. Write the polynomial made of its terms up to $x^{4}$.

Answer:

22. Somewhere new

A power series solution of $(x - 3)(x + 4)y'' + y' + y = 0$ is sought about $x = 0$. On which interval does the theorem guarantee it converges?

This task has no paper form; do it on a device.

23. Somewhere new

Near the bright edge of a rainbow the light's intensity follows Airy's equation $y'' = xy$. With $y(0) = 2$ and $y'(0) = 0$, estimate $y(0.6)$ from the series terms up to $x^{3}$.

Answer:

24. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

25. Test question

Find the power series solution of $y'' - xy' - y = 0$ about $x = 0$ with $y(0) = 16$ and $y'(0) = 0$. Write the polynomial made of its terms up to $x^{4}$.

Answer:

26. What you can do now

You can turn a variable-coefficient equation into a recurrence for the coefficients of a series, and you can state how far that series is guaranteed to converge. Say in your own words why the point must be checked before the series is written.

Working for the steps left to you

15. Your turn: find the recurrence for $y'' - 2xy' - 2y = 0$ about $x = 0$, step 2

15. Your turn: find the recurrence for $y'' - 2xy' - 2y = 0$ about $x = 0$, step 3