Hej,
I have a question about the following problem:
Derive a recurrence formula for $m \ge 0$ given the generating function formula
$$
\frac{1}{1-2xt+t^{2}}=\sum_{m=0}^\infty U_m(x)t^{m}.
$$
What I did was first taking the derivative on both sides with respect to $t$ resulting in
$$
\frac{2x-2t}{(1-2xt+t^{2})^2}=\sum_{m=0}^\infty m U_m(x)t^{m-1}.
$$
Multiplying both sides with $1-2xt+t^2$ and using the generating formula results in
$$
(2x-2t)\sum_{m=0}^\infty U_m(x)t^{m}=\left( \sum_{m=0}^\infty m U_m(x)t^{m-1}\right)(1-2xt+t^2).
$$
This can be written as
$$
0=\sum_{m=0}^\infty m U_m t^{m-1} -2x\sum_{m=0}^\infty (m+1)U_mt^m + \sum_{m=0}^\infty (m+2) U_m t^{m+1}.
$$
We can shift the first sum
$$
\sum_{m=0}^\infty mU_mt^{m-1}= \sum_{m=0}^\infty(m+1)U_{m+1}t^m.
$$
The second one is already of the same form.
Now I want to shift the third sum to get it on the same form but when I do that I get a weird term that is left over and I don't know what to do with it:
$$
\sum_{m=0}^\infty (m+2) U_m t^{m+1} = \sum_{m=0}^\infty \left((m+1) U_{m-1} t^m\right)-U_{-1}.
$$
If it wasn't for that left-over term I could just put the sums together and get the recurrence relation
$$
U_{m-1}-2xU_m+U_{m+1}=0
$$
which is what I was looking for.
Did I make a mistake in shifting the sum or is there a way of getting rid of $U_{-1}(x)$?
Best Answer
As you know, the Chebyshev Polynomial of Second Kind satisfies the relations$$U_0(x)=1$$ $$U_1(x)=2x$$ $$U_{n+1}(x)=2xU_n(x)-U_{n-1}(x)$$ Plugging $n=0$ gives $U_{-1}(x)=0$, so I think we can drop $U_{-1}$ as $0$, giving the desired conclusion.