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Define a sequence of polynomials: $p_0(x)=1$, $p_1(x)=2+x$, and, for $n\ge 1$, $p_{n+1}=(4n+2)p_n(x)+x^2p_{n-1}(x)$ so that the first few are $1, x+2, x^2+6x+12, x^2+12x^2+60x+120$.

Is there an elementary explanation for why $p_n(x)-e^xp_n(-x)=O(x^{2n+1})$?

I understand that the polynomials $\{p_n\}$ form a variant of the ``reverse Bessel polynomials”, and that the result I seek is tantamount to the fact that $p_n(x)/p_n(-x)$ is the (n,n) Pade approximation of $e^x$. I wish to avoid, for expository reasons, all the theories of continued fractions, orthogonal polynomials, Pade approximations, etc. I wouldn’t mind using the fact that $f_n(x):=e^{-x/2}p_n(x)$ satisfies the differential equation $4xf_n’’-8nf_n’-xf_n=0$ if that turns out to be helpful.

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  • $\begingroup$ This isn't true. $p_1(x)-e^xp_1(-x)$ tends to infinity at rate even faster than $e^x$, much faster than claimed $O(x^3)$. $\endgroup$
    – Wojowu
    Jul 4 '16 at 16:50
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    $\begingroup$ I think the OP means $O(x^{2n+1})$ as x tends to 0, not infinity. $\endgroup$ Jul 4 '16 at 17:05
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We can indeed deduce this from the ODE you gave. We want to show that $$ g(x):=f_n(x)-f_n(-x) = O(x^{2n+1}) $$ near $x=0$. First of all, observe that $h(x):=f_n(-x)$ satisfies the same ODE, and thus so does $g=f_n-h$. The equation has a regular singular point at $x=0$, and can be discussed with standard methods: Look for solutions of the form $g=\sum_{j\ge 0} a_j x^{j+\lambda}$. The indicial equation is $$ 4\lambda (\lambda-1) - 8n\lambda = 0 , $$ which has the solutions $\lambda_1=2n+1$, $\lambda_2=0$. So the general solution is of the form $AP(x) + B x^{2n+1}Q(x)$, where $P,Q$ are power series with $P(0),Q(0)\not= 0$. Since $g(0)=0$, the solution we're interested in is of the form $g=Bx^{2n+1}Q(x)$, as claimed.

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