Question about the exponential generating function for $T_n=T_{n-1}+(n-1)\cdot T_{n-2}$

Question

So I came across this step when finding an expontential generating function using this recurrence relation, $$ T_{n}=T_{n-1}+(n-1) \cdot T_{n-2} $$ Since $T_{0}=T_{1}=1$ For $n=0$ $$ \sum_{n \geq 0} \frac{T_{0} x^{0}}{0 !}=1 $$ For $n=1$ $$ \sum_{n \geq 1} \frac{T_{1} x^{1}}{1 !}=x $$ Therefore, $$ f(x)=\sum_{n \geq 0} \frac{T_{n} x^{n}}{n !}=1+x+\sum_{n \geq 2} \frac{T_{n} x^{n}}{n !} $$ From the recurrence relation, $T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}$. Hence, $$ \Rightarrow 1+x+\sum_{n \geq 2} \frac{T_{n-1}+(n-1) T_{n-2}}{n !} x^{n} $$ Distributing the terms using the commutative law of addition, we get, $$ \begin{gathered} \therefore 1+x+\sum_{n \geq 2} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n} \\ \Rightarrow 1+\sum_{n \geq 1} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n} \end{gathered} $$ Where did the ' $x$ ' go in the last step above? And how did the lower limit change from $n \geq 2$ to $n \geq 1$ ?

Could anyone help explain? Any help would be highly appreciated.

Answer 1

Motivation

$2$ can be written as $1 + 1$, or in fact, in any number of other ways, right?

Idea

Great!

Now that we are motivated, let us take a closer look at $$1 + x + \sum_{n \geq 2} \frac {T_{n - 1}} {n!} x^n + \sum_{n \geq 2} \frac {(n - 1) T_{n - 2}} {n!} x^n.$$ In particular, zoom in on $$\sum_{n \geq 2} \frac {T_{n - 1}} {n!} x^n.$$ Although the index starts from $n = 2$, there is no harm in trying to see what happens when $n = 1$.

When $n = 1$, observe that $$\frac {T_{n-1}} {n!} x^n = \frac {T_{1 - 1}} {1!} x^1 = T_{0} x = x.$$ This is why the '$x$' (i.e. second term) in $$1 + x + \sum_{n \geq 2} \frac {T_{n - 1}} {n!} x^n + \sum_{n \geq 2} \frac {(n - 1) T_{n - 2}} {n!} x^n$$ has seemingly "disappeared". It has simply been re-written and included into the first summation (i.e. third term).

Answer 2

Observing that $T_1=T_0=1$, now $$x=\frac{T_1 x}{1!}=\frac{T_0 x}{1!}$$ so you can absorb this term as $n=1$ of your second last summand.

Answer 3

Just compare the two lines, focusing on

$$\sum_{n\geq 1} \frac{T_{n-1}}{n!}x^n = \color{red}{\underbrace{\frac{T_{1-1}}{1!}x^1}_{n=1}} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$

$$x + \sum_{n\geq 2} \frac{T_{n-1}}{n!}x^n = \color{red}{x} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$

so it all comes down to $$\frac{T_{1-1}}{1!}x^1 \stackrel{?}{=} x$$ and this is true since $T_0 = 1$.