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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 $x - \frac{x^3}{3!} + \frac{x^5}{5!}$?

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    http://en.wikipedia.org/wiki/Analytic_function might help2012-11-30
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    It is the second.2012-11-30
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    Somewhat related: http://math.stackexchange.com/q/201523/12422012-11-30

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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.