Taylor's Theorem: The Remainder Envelope

Every derivative of $e^x$ is $e^x$, so $f^{(k)}(0) = 1$. Hence $T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$ and $T_3(1) = 1 + 1 + \frac12 + \frac16 = \frac83 \approx 2.6667$.

By Taylor's theorem (Lagrange remainder), $R_3(1) = \frac{e^c}{4!} \cdot 1^4$ for some $c \in (0, 1)$. Since $e^c \le e$, $|R_3(1)| \le \frac{e}{24} \approx 0.1133$.

The true error is $e - \frac83 \approx 0.0516 < 0.1133$, inside the bound — a remainder bound is a safe overestimate.

Every derivative of $\sin$ is $\pm\sin$ or $\pm\cos$, so $|f^{(n+1)}(c)| \le 1$. The Lagrange remainder is $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$, hence $|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!}$.

The case $x = 0$ is trivial and separate: the series is identically $0 = \sin 0$ (here $a_n = 0$ identically, so the ratio below would be $0/0$).

Assume $x \ne 0$ and put $a_n = \frac{|x|^{n+1}}{(n+1)!}$; then $\frac{a_{n+1}}{a_n} = \frac{|x|}{n+2}$.

Once $n + 2 > 2|x|$ we get $a_{n+1} \le \frac{1}{2}a_n$, so for $n \ge N := \lceil 2|x| \rceil$, $a_n \le a_N\left(\frac{1}{2}\right)^{n-N} \to 0$ — the factorial beats the power geometrically.

Hence $|T_n(x) - \sin x| \to 0$: the Maclaurin series converges to $\sin x$ for every $x$.

Assume $f(x) = \sum_{k=0}^{\infty} c_k (x-a)^k$ converges near $a$. By the power series theorem, a power series is termwise differentiable to all orders inside its radius of convergence, so the repeated termwise differentiation used below (differentiating $j$ times for arbitrary $j$) is justified — this theorem is cited here to align the hypotheses.

Differentiating $j$ times gives $f^{(j)}(x) = \sum_{k=j}^{\infty} k(k-1)\cdots(k-j+1)\,c_k\,(x-a)^{k-j}$ (terms with $k < j$ vanish in the differentiation).

Setting $x = a$ kills every term with $k > j$ via $(x-a)^{k-j}$; only $k = j$ survives, giving $f^{(j)}(a) = j!\,c_j$, i.e. $c_j = \frac{f^{(j)}(a)}{j!}$ — the Taylor coefficients are unique.