Equation with sums of functions, justification for considering only individual functions

Question

I'm not sure how to title this question, so please edit the title if you know better.

I have an equation involving sums of functions: $$ -\sum_{j=i+1}^NQ_jC_{i,j}e^{-Q_jt}=-Q_i\sum_{j=i+1}^NC_{i,j}e^{-Q_jt}+f\sum_{j=i+1}^{N}\sum_{k=i+1}^jC_{k,j}e^{-Q_jt}$$ where each $Q_j$ is unique. I want to solve for the coeffiecients $C_{i,j}$. I know I can produce a solution by considering each exponential function separately i.e. dropping the summation on j: $$ -Q_jC_{i,j}e^{-Q_jt}=-Q_iC_{i,j}e^{-Q_jt}+f\sum_{k=i+1}^jC_{k,j}e^{-Q_jt}$$

I am struggling to articulate the justification for this. Can someone explain why and when I can or can't do this? I'm specifically looking for language to use in my justification. I feel like this is sensible, but just don't know how to express why.

Answer 1

Playing around, we can convert $-\sum_{j=i+1}^NQ_jC_{i,j}e^{-Q_jt}=-Q_i\sum_{j=i+1}^NC_{i,j}e^{-Q_jt}+f\sum_{j=i+1}^{N}\sum_{k=i+1}^jC_{k,j}e^{-Q_jt} $ to, assuming that is supposed to be $e^{-Q_jt}$ and not $e^{-Q_it}$ in the rightmost sum,

$\begin{array}\\ 0 &=\sum_{j=i+1}^NQ_jC_{i,j}e^{-Q_jt}-Q_i\sum_{j=i+1}^NC_{i,j}e^{-Q_jt}+f\sum_{j=i+1}^{N}e^{-Q_jt}\sum_{k=i+1}^jC_{k,j}\\ &=\sum_{j=i+1}^N \left(Q_jC_{i,j}e^{-Q_jt}-Q_iC_{i,j}e^{-Q_jt}+fe^{-Q_jt}\sum_{k=i+1}^jC_{k,j}\right)\\ &=\sum_{j=i+1}^N \left((Q_j-Q_i)C_{i,j}e^{-Q_jt}+fe^{-Q_jt}\sum_{k=i+1}^jC_{k,j}\right)\\ &=\sum_{j=i+1}^N e^{-Q_jt}\left((Q_j-Q_i)C_{i,j}+f\sum_{k=i+1}^jC_{k,j}\right)\\ &=e^{-Q_it}\sum_{j=i+1}^N e^{-(Q_j-Q_i)t}\left((Q_j-Q_i)C_{i,j}+f\sum_{k=i+1}^jC_{k,j}\right)\\ &=\sum_{j=i+1}^N e^{-(Q_j-Q_i)t}\left((Q_j-Q_i)C_{i,j}+f\sum_{k=i+1}^jC_{k,j}\right) \qquad\text{ since } e^{-Q_it} \ne 0\\ \end{array} $

A solution for this could certainly be gotten by setting $C_{i,j} =\frac{f}{Q_i-Q_j}\sum_{k=i+1}^jC_{k,j} $, though I am not sure how to start things off for the initial values of $C_{i, j}$. If $j = i+1$, we have $C_{i,i+1} =\frac{f}{Q_i-Q_j}C_{i+1,i+1} $.

Anyway, that's all I can see here. Hope it helps.