How has this Chebyshev expansion been reindexed? Just a quick question, I'm going through my lecture notes and I can't see how the author has gone from this:
$$\begin{aligned} f ( x ) g ( x ) & = \sum _ { m = 0 } ^ { \infty } \breve { f } _ { m } T _ { m } ( x ) \sum _ { n = 0 } ^ { \infty } \breve { g } _ { n } T _ { n } ( x ) \\ & = \frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n } \left[ T _ { | m - n | } ( x ) + T _ { m + n } ( x ) \right] \end{aligned}$$
which is fine (the identity $T _ { m } ( x ) T _ { n } ( x ) =\frac { 1 } { 2 } \left[ T _ { | m - n | } ( x ) + T _ { m + n } ( x ) \right]$ is separately derived), to this:
$$= \frac { 1 } { 2 } \sum _ { n = 0 } ^ { \infty } \sum _ { m = 0 } ^ { \infty } \breve { f } _ { m } \left( \breve { g } _ { | m - n | } + \breve { g } _ { m + n } \right) T _ { n } ( x )$$
it seems like it's just a reindex and apparently it's "elementary algebra" but I can't seem to spot it. 

For context $T_n(x)$ is the $n$th Chebyshev polynomial and $f$ and $g$ have expansions of the form $$f ( x ) = \sum _ { n = 0 } ^ { \infty } \breve { f } _ { n } T _ { n } ( x )$$
 A: Extend $T_j$ to negative $j$ by $T_{-j} = T_j$ so that $T_{|j|} = T_j$. And let $ \breve { f } _ { m }= \breve { g } _ { n }=0$ for $n,m<0$. Then, 
$$\begin{eqnarray}
 \sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n }T _ { | m - n | } ( x )& =& \sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n }T _ { m - n  } ( x )\\ &=& \sum _ { m = 0 } ^ { \infty } \sum _ { n\le m} ^ { } \breve { f } _ { m } \breve { g } _ {m- n }T _ {n } ( x )\quad (m-n\mapsto n)\\
&=&\sum _ { m = 0 } ^ { \infty } \sum _ { n\ge -m} ^ { } \breve { f } _ { m } \breve { g } _ {m+ n }T _ {n } ( x )\quad (n\mapsto -n)\\
&=&\sum _ { m = 0 } ^ { \infty } \sum _ { n\ge 0} ^ { } \breve { f } _ { m } \breve { g } _ {m+ n }T _ {n } ( x )+\sum _ { m = 0 } ^ { \infty } \sum _ { n=-m} ^ { -1} \breve { f } _ { m } \breve { g } _ {m+ n }T _ {n } ( x )\\
&=&\sum _ { m = 0 } ^ { \infty } \sum _ { n\ge 0} ^ { } \breve { f } _ { m } \breve { g } _ {m+ n }T _ {n } ( x )+\sum _ { m = 0 } ^ { \infty } \sum _ { n=1} ^ { m} \breve { f } _ { m } \breve { g } _ {m-n }T _ {n } ( x )\\&=&\sum _ { m = 0 } ^ { \infty } \sum _ { n\ge 0} ^ { } \breve { f } _ { m } \breve { g } _ {m+ n }T _ {n } ( x )+\sum _ { m = 0 } ^ { \infty } \sum _ { n=1} ^ { m} \breve { f } _ { m } \breve { g } _ {|n-m| }T _ {n } ( x ).\end{eqnarray}$$ On the other hand,
$$
\sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n }  T _ { m + n } ( x )=\sum _ { m = 0 } ^ { \infty } \sum _ { n\ge m }  \breve { f } _ { m } \breve { g } _ { n-m }  T _ { n } ( x )=\sum _ { m = 0 } ^ { \infty } \sum _ { n= m }^\infty  \breve { f } _ { m } \breve { g } _ { |n-m| }  T _ { n } ( x )
.
$$ Gathering them together, we have
$$\begin{eqnarray}
\frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n } \left[ T _ { | m - n | } ( x ) + T _ { m + n } ( x ) \right]&=&\frac { 1 } { 2 }\sum _ { m = 0 } ^ { \infty } \sum _ { n=0} ^ {\infty } \breve { f } _ { m } (\breve { g } _ { |n-m| } +\breve { g } _ {m+ n })T _ {n } ( x )\\&&+\frac { 1 } { 2 }\sum_{m=0}^\infty\left(\breve { f } _ { m }\breve { g } _ {0 }T_m(x)-\breve { f } _ { m }\breve { g } _ {m }T_0(x)\right).
\end{eqnarray}$$
A: I've taken another look and I'm pretty sure this must just be incorrect for two reasons:

(1) Trying to show directly:
If we start from $$ \frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n } \left[ T _ { | m - n | }  + T _ { m + n } \right]$$
we can split the absolute value up and reaarange the terms to get
$$\frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty }\breve { f } _ { m }\left[   \sum_{n=0}^m \breve{ g } _ { n }  T _ {  m - n  }+ \sum_{n=m+1}^m \breve{ g } _ { n }  T _ {  n-m  }+\sum_{n=0}^\infty \breve { g } _ { n }  T _ { m+n } \right]$$
then we can reindex each sum in the square brackets according to the rules: $n\to n-m$ for the first, $n\to m-n$ for the second and $n\to n+m$ for the third. Which results in:
$$ \frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty }\breve { f } _ { m }\left[ \sum_{n=m}^\infty \breve{g}_{n-m}T_n+\sum_{n=0}^m \breve{g}_{m-n}T_n+\sum_{n=1}^\infty \breve{g}_{n+m}T_n  \right] $$
which I can force into the form given by writing this as:
$$ \frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty }\breve { f } _ { m }\left[ \sum_{n=0}^\infty (\breve{g}_{|n-m|}+\breve{g}_{n+m})T_n +\breve{g}_0 T_m-\breve{g}_mT_0 \right] $$
where the it only differs by those two terms. 

(2) Check equality for $T_0$ terms
First let's look at the formula $$\frac { 1 } { 2 } \sum _ { m = 0 } ^ { \infty } \sum _ { n = 0 } ^ { \infty } \breve { f } _ { m } \breve { g } _ { n } \left[ T _ { | m - n | } + T _ { m + n } \right]$$
to get the terms involving $T_0$ consider


*

*Fixed $m>0$, we need $n=m$ whose terms are: $\breve{f}_m\breve{g}_m (T_0)$

*$m=0$, we need $n=m=0$ whose term is: $\breve{f}_0\breve{g}_0(T_0+T_0)=2\breve{f}_0\breve{g}_0T_0$
This is different for the formula $$\frac { 1 } { 2 } \sum _ { n = 0 } ^ { \infty } \sum _ { m = 0 } ^ { \infty } \breve { f } _ { m } \left( \breve { g } _ { | m - n | } + \breve { g } _ { m + n } \right) T _ { n } ( x )$$
since if we repeat the procedure we get:


*

*Fixed $m>0$, we need $n=0$ whose terms are: $\breve{f}_m(\breve{g}_m+\breve{g}_m) T_0=2\breve{f}_m\breve{g}_m T_0$

*$m=0$, we need $n=m=0$ with term: $\breve{f}_0(\breve{g}_0+\breve{g}_0)T_0=2\breve{f}_0\breve{g}_0T_0$
which would incidentally be fixed with the extra $-\breve{g}_m T_0$ from the direct way above.

edit: (3) Another reason
Just from implementing in Mathematica with $f(x)=\sin(x)$ and $g(x)=\cos(x)$  then the one they give (labeled A) vs. the one I derived (labeled B):


This result is in two published books so I feel like I have made a mistake somewhere so if anyone spots anything please let me know.
