Ask Uncle Colin: about Chebyshev’s Equation

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

Dear Uncle Colin,

I've been asked to solve Chebyshev's equation using a series expansion:

$(1-x)\diffn{2}{y}{x} - x\dydx + p^2 y = 0$

assuming $y=C_0 + C_1 x + C_2 x^2 + ...$.

I end up with the relation $C_{N+2} = \frac{C_N \left(N^2 -p^2\right)}{(N+2)(N+1)}$, but the given answer has a + on top. Where have I gone wrong?

- Can't Have Everything, But Your Signs Have Erroneously Varied

Hi, CHEBYSHEV, and thanks for your message!

The TL;DR here is, I think your answer is correct and the given answer wrong. I presume this is from a Physics course, where the odd minus sign is seen as an occupational hazard.

The trickiest part (for me) with series solutions is keeping the indices and subscripts straight, so I generally write down a few terms for each of the parts to make sure I have them right before doing anything clever with $\Sigma$s.

Here, I'd start with the various derivatives:

  • $y=C_0 + C_1 x + C_2 x^2 + ... + C_n x^n + ...$
  • $y' = C_1 + 2C_2 x + 3 C_3 x^2 + ... + n C_n x^{n-1} + ...$
  • $y'' = 2C_2 + 6C_3 x + 12 C_4 x^2 + ... + n(n-1) C_n x^{n-2} + ...$

And then look at the four (yes, four) terms of the differential equation:

  • $y'' = \sum_{n=2}^\infty C_n n(n-1) x^{n-2}$
  • $-x^2 y'' = \sum_{n=2}^\infty -C_n n(n-1) x^n$
  • $-xy' = \sum_{n=1}^\infty -n C_n x^n$
  • $p^2 y = \sum_{n=0}^\infty C_n x^n p^2$

All of those have $x^n$ terms in except for the first one, which has $x^{n-2}$. If I change the variable so $N = n-2$, it becomes:

$y'' = \sum_{N=0}^\infty C_{N+2} (N+2)(N+1) x^N$

And we want the coefficients for each power of $x$ to sum to 0. (With a little sleight of hand, I'm going to turn those placeholder $N$s back into $n$s.)

That gives us:

$C_{n+2} (n+2)(n+1) - C_n (n-1)n - n C_n + C_n p^2 = 0$, for $n \ge 2$

Rearranging, $C_{n+2} = C_n \frac{ n(n-1)+n - p^2}{(n+2)(n+1)}$, which simplifies to

$C_{n+2} = C_n \frac{n^2 - p^2}{(n+2)(n+1)}$, as you have.

As a belt-and-braces approach, I also looked it up on MathWorld, which agrees with our answers.

Physicists, eh?

Hope that helps,

- Uncle Colin


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


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