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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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en.wikipedia.org/wiki/Analytic_function might help –  sidht Nov 30 '12 at 19:54
    
It is the second. –  André Nicolas Nov 30 '12 at 19:55
    
Somewhat related: math.stackexchange.com/q/201523/1242 –  Hans Lundmark Nov 30 '12 at 21:13
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up vote 2 down vote accepted

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.

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