Obtaining a recursion relation

Question

I have obtained an equation using the power series solution method. For the sake of simplicity, I have substituted the complicated constants by A,B,C,... So the general form of my equation is $$ A\sum_{i=0}^n a_i\,x^{i+3}-\sum_{i=0}^n\,[B-(i+1)]a_i\,x^{i+1}-\sum_{i=0}^na_i\,x^i-\sum_{i=0}^n\frac{i(i+2)}{C}\,a_i\,x^{i-1}=0 $$ How should I proceed to get an equation including the same powers? and then get recursion relation?

Answer 1

Starting from what you wrote $$A\sum_{i=0}^n a_i\,x^{i+3}-\sum_{i=0}^n\,[B-(i+1)]a_i\,x^{i+1}-\sum_{i=0}^na_i\,x^i-\sum_{i=0}^n\frac{i(i+2)}{C}\,a_i\,x^{i-1}=0$$ consider what it will be for the term $x^k$ (that is to say that in each sum you fix the index $i$ such that the exponent of $x$ is $k$). Doing it, $$A a_{k-3}-[B-k]a_{k-1}-a_k-\frac{(k+1)(k+3)}{C}\,a_{k+1}=0$$

Answer 2

Note that we can write \begin{multline*} 0 = A\sum_{i=0}^n a_ix^{i+3}-\sum_{i=0}^n[B-(i+1)]a_ix^{i+1}-\sum_{i=0}^na_ix^i-\sum_{i=0}^n\frac{i(i+2)}{C}a_ix^{i-1} \\ = A\sum_{i=3}^{n+3} a_{i-3}x^{i}-\sum_{i=1}^{n+1}[B-i]a_{i-1}x^{i}-\sum_{i=0}^na_ix^i-\sum_{i=-1}^{n-1}\frac{(i+1)(i+3)}{C}a_{i+1}x^{i}. \end{multline*} What is the coefficient of $x^i$ on the right-hand side? (Perhaps restrict to $3\le i\le n-1$ at first.) What is the coefficient of $x^i$ on the left-hand side?