Ytukyg

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 $$f ( x ) = sum _ { n = 0 } ^ { infty } breve { f } _ { n } T _ { n } ( x )$$

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,
$$begin{eqnarray}
sum _ { m = 0 } ^ { infty } sum _ { n = 0 } ^ { infty } breve { f } _ { m } breve { g } _ { n }T _ { | m - n | } ( x )& =& sum _ { m = 0 } ^ { infty } sum _ { n = 0 } ^ { infty } breve { f } _ { m } breve { g } _ { n }T _ { m - n } ( x )\ &=& sum _ { m = 0 } ^ { infty } sum _ { nle m} ^ { } breve { f } _ { m } breve { g } _ {m- n }T _ {n } ( x )quad (m-nmapsto n)\
&=&sum _ { m = 0 } ^ { infty } sum _ { nge -m} ^ { } breve { f } _ { m } breve { g } _ {m+ n }T _ {n } ( x )quad (nmapsto -n)\
&=&sum _ { m = 0 } ^ { infty } sum _ { nge 0} ^ { } breve { f } _ { m } breve { g } _ {m+ n }T _ {n } ( x )+sum _ { m = 0 } ^ { infty } sum _ { n=-m} ^ { -1} breve { f } _ { m } breve { g } _ {m+ n }T _ {n } ( x )\
&=&sum _ { m = 0 } ^ { infty } sum _ { nge 0} ^ { } breve { f } _ { m } breve { g } _ {m+ n }T _ {n } ( x )+sum _ { m = 0 } ^ { infty } sum _ { n=1} ^ { m} breve { f } _ { m } breve { g } _ {m-n }T _ {n } ( x )\&=&sum _ { m = 0 } ^ { infty } sum _ { nge 0} ^ { } breve { f } _ { m } breve { g } _ {m+ n }T _ {n } ( x )+sum _ { m = 0 } ^ { infty } sum _ { n=1} ^ { m} breve { f } _ { m } breve { g } _ {|n-m| }T _ {n } ( x ).end{eqnarray}$$ On the other hand,
$$
sum _ { m = 0 } ^ { infty } sum _ { n = 0 } ^ { infty } breve { f } _ { m } breve { g } _ { n } T _ { m + n } ( x )=sum _ { m = 0 } ^ { infty } sum _ { nge m } breve { f } _ { m } breve { g } _ { n-m } T _ { n } ( x )=sum _ { m = 0 } ^ { infty } sum _ { n= m }^infty breve { f } _ { m } breve { g } _ { |n-m| } T _ { n } ( x )
.
$$ Gathering them together, we have
$$begin{eqnarray}
frac { 1 } { 2 } sum _ { m = 0 } ^ { infty } sum _ { n = 0 } ^ { infty } breve { f } _ { m } breve { g } _ { n } left[ T _ { | m - n | } ( x ) + T _ { m + n } ( x ) right]&=&frac { 1 } { 2 }sum _ { m = 0 } ^ { infty } sum _ { n=0} ^ {infty } breve { f } _ { m } (breve { g } _ { |n-m| } +breve { g } _ {m+ n })T _ {n } ( x )\&&+frac { 1 } { 2 }sum_{m=0}^inftyleft(breve { f } _ { m }breve { g } _ {0 }T_m(x)-breve { f } _ { m }breve { g } _ {m }T_0(x)right).
end{eqnarray}$$

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: $nto n-m$ for the first, $nto m-n$ for the second and $nto n+m$ 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: $breve{f}_mbreve{g}_m (T_0)$
  


• 
  $m=0$, we need $n=m=0$ whose term is: $breve{f}_0breve{g}_0(T_0+T_0)=2breve{f}_0breve{g}_0T_0$
  

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: $breve{f}_m(breve{g}_m+breve{g}_m) T_0=2breve{f}_mbreve{g}_m T_0$
  


• 
  $m=0$, we need $n=m=0$ with term: $breve{f}_0(breve{g}_0+breve{g}_0)T_0=2breve{f}_0breve{g}_0T_0$
  

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.