Taylor polynomial and degree

Question

I read that one can form Taylor polynomials for some functions, like $$\sin x\approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}.$$ Is it correct to say that $\sin x$ has no Taylor polynomial with center 0 of order six or is the sixth order Taylor polynomial actually $x - \frac{x^3}{3!} + \frac{x^5}{5!}$?

Answer 1

As you say, it is actually the same as that of order $5$. Note that if $T_{n,a}$ is the Taylor polynomial of order $n$ about $a$ of $f$, then $\deg(T)\leq n$, as it happens in this case.