Taylor polynomial related series

Question

Given $f:\mathbb{R} \to \mathbb{R}$ an analytical function, $$ (T_nf)(x) = \sum_{k=0}^{n} {\frac{f^{(k)}(0)}{k!}x^k} $$ and $$ (R_nf)(x) = f(x) -(T_nf)(x), $$ calculate the sum of the following series: $$ \sum_{n=0}^{\infty} {(R_nf)(t)x^n}$$ where $t, x \in \mathbb{R}$

Answer 1

As far as I'm concerned, this is a clear case of a technique known as "let's just unroll the definitions and see where that leads us, shall we?"

• For $t\in\mathbb{R}$ and $x\neq 1$, letting $a_k \stackrel{\rm def}{=} \frac{f^{(k)}(0)}{k!}$ for all $k\geq 0$, $$ \begin{align} \sum_{n=0}^\infty (R_n f)(t) x^n &= \sum_{n=0}^\infty \sum_{m=n+1}^\infty a_m t^m x^n = \sum_{m=0}^\infty \sum_{n=0}^{m-1} a_m t^m x^n = \sum_{m=0}^\infty a_m t^m \sum_{n=0}^{m-1} x^n \\ &= \sum_{m=0}^\infty a_m t^m \frac{1-x^{m}}{1-x} = \frac{1}{1-x} \left( \sum_{m=0}^\infty a_m t^m - \sum_{m=0}^\infty a_m (xt)^m \right)\\ &= \frac{f(t) - f(xt)}{1-x} \end{align}$$

For $x=0$: we do get $R_0 f(t) = f(t)-f(0)$, so this passes a simple sanity check.

• For $x=1$, you want to compute $\sum_{m=0}^\infty m a_m t^m = t \sum_{m=1}^\infty m a_m t^{m-1} = t f'(t)$.