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when solving for the coefficients for the Legendre ODE $(1-x^2)y’’-2xy’+l(l+1)y=0$, I understand how to obtain the recurrence relation

$$a_{k+2}=\frac{k(k+1)-l(l+1)}{(k+2)(k+1)}a_k.$$

What I do not understand is the distinction between infinite series and terminating series. I know that for both $k$ and $l$ even/odd the series “terminates” to the Legendre polynomials, while other combinations result in an “infinite series”.

I do not get the meanings of the quoted terms here. For one, isn’t part of the solution the sum of $c_l P_l$, which ranges from $l=0$ to $l=\infty$, making that an infinite series? Likewise, doesn’t the other solution sum up nicely into a natural logarithm?

Additionally, how do I extend this concept to any other type of ODE?

Apologies for the buffoonish question here, I am incredibly confused!

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  • $\begingroup$ @Somos sorry, I don’t really understand what you’re trying to get at $\endgroup$
    – user107224
    Jan 1, 2019 at 13:48
  • $\begingroup$ @Somos zero, but this still doesn’t give me clear definitions of terminated vs infinite $\endgroup$
    – user107224
    Jan 1, 2019 at 14:53
  • $\begingroup$ @Somos doesn’t the same thing happen for all combinations of even and odd k and l? I don’t see his setting $k=l$ leads to the polynomial series too. Additionally, this seems to be specific to the Legendre equation, what about for a general ODE? $\endgroup$
    – user107224
    Jan 1, 2019 at 15:20

1 Answer 1

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If the terms of an infinite series $\,s_1 + s_2 + ... + s_n + ...\,$ are such that they are equal to zero after $\,s_n\,$, then it is said to terminate and its sum is $\,s_1 + s_2 + ... + s_n\,$ which is a finite sum and the series converges to it.

In the common case of a power series $\,a_0 + a_1 x + a_2 x^2 + ... + a_n x^n + ...\,$ the same thing applies and a terminating power series is $\,a_0 + a_1 x + a_2 x^2 + ... + a_nx^n\,$ which is a polynomial and which has infinite radius of convergence.

You can look at MSE question 2573694 "Validity of terminating series solution of differential equation" for a similar situation.

Note that I think that the terminology is not a good one, but it is commonly used -- likely because of a lack of a better one.

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