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The function is $$f(x) = 2\cos x. $$ Again, $a = \pi/2$ and $n \to\infty$.

I was to do this for $n = 3$ as well and I had no problems doing that at all, but I'm confused on how you do it for for $n \to\infty$. Any tips for getting started on that?

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Generally, it's good to make your question self-contained; that includes not distributing it between the title and the main text, specifying the context and defining variables you use. In this case, it's easy enough to guess that you're probably talking about a Taylor expansion, that $a$ is the point around which you're expanding and $n$ is the order of the expansion, but preferably we shouldn't have to guess. –  joriki May 24 '11 at 7:46
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More specifically, it's not clear (to me) what you mean by "finding Lagrange's form of the remainder with $a=\pi/2$ and $n\to\infty$" -- the remainder is different for each $n$; the only thing you can say about it for $n\to\infty$ is that it goes to zero. If it's the limiting behaviour you're interested in, you should state that more explicitly. –  joriki May 24 '11 at 7:48
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@joriki: I'm sorry if it was confusing, I really did just read it off the paper though. It just states the function (2cosx) and then says to find Langrange's form of the remainder R(x) with a=π/2 and with n-> ∞. It also asked to find it with n = 3 and all I had to do for that one was plug in stuff to the formula and then take a bunch of derivatives. –  Ryan May 24 '11 at 8:07
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In that case, it's a poorly worded exercise -- I guess they just mean you should find the limit of the remainder as $n$ goes to infinity. To do that, you can look at the relative growth of the power term and the factorial. –  joriki May 24 '11 at 8:48
    
What are $a$ and $n$ in your question? –  Jack Jun 23 '11 at 17:32

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Hint: For any $x \in \mathbb{R}$ fixed, and as $n \to \infty$, $$ |R_n (x)| \le \frac{{2|x - a|^{n + 1} }}{{(n + 1)!}} \to 0. $$

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