Equalizing of powers in power series solution method

Question

I have reached to the following equation using power series solution method $$ \sum_{i=0}^p(i+1)\,a_i\,r^{i+1}+\sum_{i=0}^pa_i\,r^i+\sum_{i=0}^p(i+2)\,a_i\,r^{i-1}=0 $$ I really don't know what should I do to equalize the powers. Do limits of summations change during this process? and if yes, how?

Answer 1

$$\sum_{i=0}^p(i+1)\,a_i\,r^{i+1}+\sum_{i=0}^pa_i\,r^i+\sum_{i=0}^p(i+2)\,a_i\,r^{i-1}=0$$

For each term, make the power equal to $k$ to get the required $i$.

So, for the first term, $i+1=k\implies i=k-1$, for the second $i=k$ and for the third, $i-1=k\implies i=k+1$.

Rewriting $$ka_{k-1}+a_k+(k+3)a_{k+1}=0$$

Edit

I made a mistake assuming that the summations are going to $\infty$. Since, from comments, this is not the case, we must be more careful.

The terms will be $$\frac{2a_0} x+(a_0+3a_1)+\sum_{k=1}^{p-1}\big(ka_{k-1}+a_k+(k+3)a_{k+1}\big)x^k+(pa_{p-1}+a_p)x^{p}+(p+1)a_p x^{p+1}$$ But this then implies $a_p=a_{p-1}=0$ as well as $a_0=a_1=0$. Then, this reduces the problem of $$\sum_{i=1}^{p-2}(i+1)\,a_i\,r^{i+1}+\sum_{i=1}^{p-2} a_i\,r^i+\sum_{i=1}^{p-2}(i+2)\,a_i\,r^{i-1}=0$$ to which apply $$ka_{k-1}+a_k+(k+3)a_{k+1}=0$$

Answer 2

We shift indices of the series to easily collect coefficients of equal powers of $r$. We obtain for integral $p\geq 0$: \begin{align*} \sum_{i=0}^p&(i+1)a_ir^{i+1}+\sum_{i=0}^pa_ir^i+\sum_{i=0}^p(i+2)a_ir^{i-1}\\ &=\sum_{i=1}^{p+1}ia_{i-1}r^i+\sum_{i=0}^pa_ir^i+\sum_{i=-1}^{p-1}(i+3)a_{i+1}r^i=0\tag{1} \end{align*}

In (1) we shift the index of the left-most sum by $+1$ and of the right-most sum by $-1$ to obtain equal powers $r^i$.

We obtain by collecting coefficients of equal powers from (1): \begin{align*} \sum_{i=-1}^{-1}&(i+3)a_{i+1}r^i+\sum_{i=0}^0a_ir^i+\sum_{i=1}^1ia_{i-1}r^i\tag{$p=0$}\\ &=2a_0\frac{1}{r}+a_0+a_0r\\ &=0\\ \sum_{i=-1}^{0}&(i+3)a_{i+1}r^i+\sum_{i=0}^1a_ir^i+\sum_{i=1}^2ia_{i-1}r^i\tag{$p=1$}\\ &=\left(2a_0\frac{1}{r}+3a_0\right)+\left(a_0+a_1r\right)+\left(a_0r+2a_1r^2\right)\\ &=2a_0\frac{1}{r}+4a_0+(a_0+a_1)r+2a_1r^2\\ &=0\\ \sum_{i=-1}^{p-1}&(i+3)a_{i+1}r^i+\sum_{i=0}^pa_ir^i+\sum_{i=1}^{p+1}ia_{i-1}r^i\tag{$p\geq 2$}\\ &=2a_0\frac{1}{r}+3a_1+a_0+\sum_{i=1}^{p-1}\left(ia_{i-1}+a_i+(i+3)a_{i+1}\right)r^i\\ &\qquad+\left((p-1)a_{p-1}+a_{p}\right)r^p+(p+1)a_pr^{p+1}\\ &=0 \end{align*}

We observe that only the trivial solutions $a_j=0, j\geq 0$ occur for each $p\geq 0$.

In the case $p\geq 2$ we separated the terms with $i\in\{-1,0,p,p+1\}$ and collected the other terms within one sum.