Equation with sums of functions, justification for considering only individual functions 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.
 A: 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.
