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Find the $n$-th order Taylor Polynomials for $f(x)=\sin x$ centered at $0$ and at $\frac{pi}{6}$; call these $T_{0,n}(x)$ and $T_{\pi/6,n}$, respectively. Then show that the sequence $(T_{0,n}(x))$ converges uniformly to $f(x)$ on any interval $[-M, M]$.

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    @Dirk To be honest I'm stuck at the beginning. I know what the general Taylor Polynomial looks like, but what is the difference between that and the n-th order.2012-12-09

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The Taylor polynomial of $f$ at $0$ is $T_{2n+1}=\sum_{k=0}^{n}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$ By the Lagrange form of the Taylor Remainder, for $x\neq 0$, $R_{2n+1}(x)=\frac{f^{(2n+2)}(\xi)}{(2n+2)!}x^{2n+2}$ for some $\xi\in (0,\left|x\right|)$. Observe that $\left|R_{2n+1}(x)\right|\le\frac{1}{(2n+2)!}\left|x\right|^{2n+2}\to 0$ as $n\to +\infty$ and so $T_{2n+1}\to f$ pointwise.

Now you can use the fact that a convergent power series converges uniformly on closed and bounded intervals or do the following: $\left|\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right|= \frac{\left|x\right|^{2k+1}}{(2k+1)!}\le \frac{M^{2k+1}}{(2k+1)!} $ for $x\in [-M,M]$. The series $\sum_{k=0}^{\infty}\frac{M^{2k+1}}{(2k+1)!}$ converges (why?) and by the Weierstrass M-test, the convergence is uniform in $[-M,M]$