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I know that $T_n(x)$ is the solution of the differential equation $(1-x^2)y''-xy'+n^2y=0$, where $$ T_n(x)=\begin{cases} T_n(x)=1 & \text{if $n=0$}\\ T_n(x)=x & \text{if $n=1$}\\ T_{n}(x)=2xT_{n-1}(x)- T_{n-2}(x) & \text{if $n\geq 2$}\\ \end{cases} $$ this can be proved using power series (https://en.wikipedia.org/wiki/Chebyshev_equation).

I was wondering if there is a way to go "backwards", given any recurrence (as a generating function) $f_n$, can I construct a differential equation such that $f_n$ is a solution of the constructed equation?.

For example, we know that $$T_n(x)=\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n} {2},$$ How can I construct $(1-x^2)y''-xy'+n^2y=0$ given that $y(x)=\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n} {2}$ is a solution of that equation?.

Thanks in advance.

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  • $\begingroup$ It's a bit difficult to see exactly what you are asking. The case of Chebyshev polynomials is well-studied from many angles and presumable all that you want to know about it is already known. Perhaps you could give an example of a situation where you would like to apply this idea but you don't already known the answer. Also, it is confusing what you call a "generating function". As you've written it, $T_n(x)$ is not a generating function for the recurrence relation you've given. $\endgroup$
    – Igor Khavkine
    Commented Oct 10, 2022 at 12:53

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In general, it is not so. For example, orthogonal polynomials satisfy a second order linear ODE if and only if they are "classical orthogonal polynomials",

MR0826863 Duistermaat, J. J.; Grünbaum, F. A. Differential equations in the spectral parameter. Comm. Math. Phys. 103 (1986), no. 2, 177–240.

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