Why are these two problems equivalent?

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I have a problem for which I don't know how to give a "fluent" explanation of it. Here is the problem, I have the following recursion

$$ P_k(x) = \left\{ \begin{array}{lr} (2k-1)xP_{k-1}(x) + (1-x^2)\frac{d}{dx}P_{k-1}(x) & k \geq 2 \\ 1 & k = 1 \end{array} \right. $$

And from this recursion I want to express the coefficients of $P_k$ given the ones of $P_{k-1}$. I already know that for each $k$ the polynomial is at most of degree $k-1$, so for each $k$ the expression is like

$$ P_k(x) = \sum_{j=0}^{k-1}a_{j,k}x^j $$

However I preferred to tackle the problem in the following way, I've studied the "equivalent problem":

$$ U_k(x) = \left\{ \begin{array}{lr} (2k-1)xU_{k-1}(x) + (1-x^2)\frac{d}{dx}U_{k-1}(x) & k \geq 2 \\ 1 & k = 1 \end{array} \right. $$

Where

$$ U_k(x) = \sum_{j=-\infty}^{+\infty} b_{j,k}x^j $$

And I've found this problem more useful because it allows me to neglect some boundary conditions that comes out. Anyway... my question is... how can I justify that the two problems expressed in terms of $P_k$ and $U_k$ are equivalent?