Finding Taylor or Maclaurin Series for a Function BC only

A Taylor series is the polynomial of 10.11 without an upper limit, so the derivative order, the factorial and the exponent are still one number and the centre still appears twice; what is new is the need for a general term valid for every $n$, and the fact that convergence becomes a question. The sine cycle shows why the general term takes care: the derivatives of $\sin$ at $0$ run $0, 1, 0, -1$ and repeat, so half the coefficients vanish and the series index stops agreeing with the derivative order. The general term is assembled from three separable pieces, the sign $(-1)^{n}$, the exponent $2n+1$ for odd powers, and a factorial matching that exponent rather than the index, giving $\sin x = \sum \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$.

The check is mechanical and worth doing every time: substituting $n = 0, 1, 2$ returns $x$, $-\frac{x^{3}}{3!}$ and $\frac{x^{5}}{5!}$, which match the terms written out by hand, and it catches all three ingredients failing. Away from the origin the general form reasserts itself, and $\ln x$ about $a = 1$ has $f^{(n)}(1) = (-1)^{n-1}(n-1)!$, which divided by $n!$ leaves $\frac{(-1)^{n-1}}{n}$ and gives $\ln x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^{n}$. Two things are visible there: the factorials cancel rather than being absent, so a general term without one can be correct, and the sum starts at $n = 1$ because $f(1) = 0$, which is what changes the sign convention. Writing $x^{n}$ for $(x-1)^{n}$ gives a series that converges to something other than $\ln x$ near $1$.