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When does the remainder term in the taylor series go to zero?

Theorem: Let $f\in C^{N+1}([\alpha,\beta])$ and $x,x_0\in(\alpha,\beta)$. Then

$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2+...+\frac{f^{(N)}(x_0)}{N!}(x-x_0)^N+\frac{(x-x_0)^{N+1}}{N!}\int_0^1 (1-t)^Nf^{(N+1)}(x_0+t(x-x_0))dt$$

Where the remainder term, $R_N$, is

$$R_N=\frac{(x-x_0)^{N+1}}{N!}\int_0^1 (1-t)^Nf^{(N+1)}(x_0+t(x-x_0))dt$$

I'm really not understanding this concept, especially because I'm struggling to comprehend what $R_N$ actually means.

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  • $\begingroup$ I suppose when the series converges. Does that make any sense ? $\endgroup$ – Claude Leibovici Jun 10 '14 at 15:05
  • $\begingroup$ I've the remainder term converges to zero, then the entire series will converge to whatever function we are approximating. But when will the remainder term converge to zero? $\endgroup$ – Mr Croutini Jun 10 '14 at 15:06
  • $\begingroup$ I don't think that answers the question. When the remainder converges to zero, $f$ is analytic at $x_0$. But when will the remainder converge to zero? $\endgroup$ – Mr Croutini Jun 10 '14 at 15:10
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There is no easy answer to the question of how to prove that the remainder term goes to zero. It is an art. The art of bounds, the mathematical art known as "Analysis".

If you try some examples you may begin to develop a mastery at this art. For instance, try $f(x) = \cos(x)$. The absolute value of the integrand is $$\bigl| \, (1-t)^Nf^{(N+1)}(x_0+t(x-x_0)) \, \bigr| $$ This is easy to bound. Since all derivatives of $\cos(x)$ are just $\pm\sin(x)$ or $\pm\cos(x)$, and since $0 \le t \le 1$, it follows that the integrand is $\le 1$ in absolute value, for all $t$. So, integrating between $0$ and $1$, and using that $\bigl| \int (blah) \bigr| \le \int |blah|$, we get $$|R_N| \le \frac{|x-x_0|^{N+1}}{N!} $$ Since $|x-x_0|$ is independent of $N$, this comes down to a standard limit problem that you should know from calculus, namely $$\lim_{N \to \infty} a^N/N!=0 $$

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  • $\begingroup$ Excellent explanation! Thank you! $\endgroup$ – Mr Croutini Jun 10 '14 at 16:03
  • $\begingroup$ Just to note: here you can see the general fact that $|R_n| \leq {|x-x_0|}^{N+1}$ I was pointing at. Kudos, excellent explanation. $\endgroup$ – JoeyBF Jun 10 '14 at 16:09
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The idea behind Taylor series is to approximate an $N+1$-times continuously differentiable function at a point by using the information given by the derivatives at that point. For example, when a function is differentiable and its derivative is continuous, we can give a linear approximation of this function at any point. This is the first order Taylor approximation.

Of course we cannot reconstruct the whole function (say, $f$) from it's Taylor approximation of any finite order, since information about $f$ is lost. In particular, when a function is not infinitely differentiable, we only have the Taylor approximation available up to a certain order (namely, $N$). The remainder represents the lost information, and can be defined precisely. However Taylor's theorem states that the higher the order of the approximation, the more negligible the remainder is, when close to the point of approximation, $x_0$. Technically, this can be stated precisely using orders of convergence.

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  • $\begingroup$ This helps my general understanding of the Taylor Series, thank you. $\endgroup$ – Mr Croutini Jun 10 '14 at 15:15
  • $\begingroup$ It is a pleasure. Feel free to ask other questions to broaden even more your understanding. $\endgroup$ – JoeyBF Jun 10 '14 at 15:17
  • $\begingroup$ Can't we just compute the remainder term? If that converges to zero, then the taylor series converges to the function we're attempting to approximate? $\endgroup$ – Mr Croutini Jun 10 '14 at 15:22
  • $\begingroup$ The Taylor formula is defined (or should I say, "Talyor"-made?) in such a way that the remainder always converges to 0 as $x \to x_0$. When adding orders of approximation however, you increase the order of convergence of the remainder. Precisely, when you have an $N$th order Taylor approximation as $x \to x_0$, the remainder converges to 0 as faster than ${|x-x_0|}^N$. Edit: Sorry, the formula was wrong. I corrected it. $\endgroup$ – JoeyBF Jun 10 '14 at 15:32
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    $\begingroup$ In our lectures we were told that the remainder term doesn't always converge to zero. Also, I like the "Taylor-made" pun! $\endgroup$ – Mr Croutini Jun 10 '14 at 15:47
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Hint: The Taylor series is sometimes called a Taylor polynomial. What is the Taylor series of a polynomial function?

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  • $\begingroup$ It's the series without the remainder term? i.e the remainder term is zero? But when does it happen? $\endgroup$ – Mr Croutini Jun 10 '14 at 15:07
  • $\begingroup$ Is it always zero? Why is the remainder term zero? $\endgroup$ – Tom R Jun 10 '14 at 15:18
  • $\begingroup$ I don't know, that's my question. When does it go to zero? $\endgroup$ – Mr Croutini Jun 10 '14 at 15:19
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    $\begingroup$ I've misunderstood your question. I thought you asked, when is the remainder term zero. The correct answer is given in the original comments. See also the wikipedia page on analytic functions. $\endgroup$ – Tom R Jun 10 '14 at 15:33
  • $\begingroup$ Okay, thank you for your help. $\endgroup$ – Mr Croutini Jun 10 '14 at 15:45

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