recurrence relation of 2nd-order

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The following is a purely computational; including a process of converting a generating function to its formal power series relations. This can be also seen as an example of recurrence relation at the 2nd-order with indices coefficient.

Observe a case of e(t)=\sum \frac{a_n}{n!}t^n=(\frac{1+t}{1-t})^{80}
To determine the a_n, notice that the k-derivative has a form

    \[\displaystyle{ e^{(k)}(t) = 160k!\frac{f^{80-k}}{(1-t)^{2k}}g_k(t) }\]

where f=(1+t)/(1-t) and g_k(t) is a monic polynomial of degree (k-1) with g_0=160^{-1}.

By a calculation, one can see a_{k+1} has an inductive formula

    \[a_{k+1}=160k!(160g_k(0)+g_k'(0))\]

And g_k has

    \[(k+1)g_{k+1}(t)= 2(tk+80)g_k(t)+(1-t^2)g_k'\]

Combining these formula, we have

    \[\displaystyle{ 160 g_k(0) = \frac{a_k}{k!} }\]

First few g_k(t) and a_k are as follows.

    \[\begin{center} \begin{tabular}{ l | l | l } \hline k $\backslash$ * & $g_k(t)$ & $a_k$ \\ \hline 0 & $160^{-1}$ & 1 \\ \hline 1 & 1 & 160 \\ \hline 2 & t+80 & 160\cdot 2!\cdot 80 \\ \hline 3 & t^2+160t+4267 & 160\cdot 3!\cdot 4267 \\ \hline 4 & t^3+240t^2+12801t+170720 & 160\cdot 4!\cdot 170720 \\ \hline \end{tabular} \end{center}\]

By close look at the first-order term of g_k, g'_{k+1}(0)=kg_k(0) holds. Thus there we have recurrence relation determined solely by means of a_k itself

    \[\begin{array}{lcl} a_{k+1}&=&160k!(\frac{a_k}{k!}+\frac{k-1}{160}\frac{a_{k-1}}{(k-1)!}) \\ &=& 160a_k+k(k-1)a_{k-1} \end{array}\]

    \[\frac{d^n}{dt^n}\big[(1-t)^{-k}(1+(1-k)t)\big]=\begin{cases} -\frac{1}{(1-t)^{k+1}} \big\{ -1+(1-k)^2t \big\} & n=1 \\ -\frac{k}{(1-t)^{k+2}} \big\{ k-3+(1-k)^2t \big\} & n=2 \\ -\frac{k(k+1)}{(1-t)^{k+3}} \big\{ 2k-5+(1-k)^2t \big\} & n=3 \end{cases}\]