# $R_{n}\left ( x \right )$ for $f\left ( x \right) = \cos\left ( 2\cdot x \right )$?

How can I define $R_{n}\left ( x \right )$ for
$$f\left ( x \right) = \cos\left ( 2\cdot x \right )$$

I found taylor expansion for $cos2x$.What should I do after that? My problem is I dont know if I should use $2n+1$ or $2n+2$ for $R_{n}\left ( x \right )$?

$$\cos\left ( 2x \right ) = \sum_{n=0}^{\infty} \frac{\left( -1 \right )^{n} \cdot \left ( 2x \right )^{2n}}{\left ( 2n \right )!}$$

And finally,what is the limit of $R_{n}\left ( x \right )$ ?

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Here is the technique. –  Mhenni Benghorbal Jul 21 '13 at 5:26

The question does not state how $R_n$ is defined. I think the most common usage is $R_n(x)=f(x)-T_n(x)$. Here $T_n$ is the Taylor polynomial, which by definition is $$T_n(x)=\sum_{k=0}^n \frac{f^{(k)}(a)}{n!}(x-a)^n$$ Therefore, for the cosine
$$R_n(x)=\sum_{k>n/2 } \frac{(-1)^{2k}}{(2k)!}(2x)^{2k}$$ The summation can also be written as $\displaystyle \sum_{k=\lfloor n/2\rfloor+1 }^\infty$.
It would be $R_{2n}(x)$, which happens to be $R_{2n+1}(x)$.
I think using $2n+1$ would be technically correct. However, using $2n+2$ often yields more powerful (i.e. useful) estimates. I suggest using that.