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How many derivatives must we take to consider some Taylor series an accurate reflection of a function?

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closed as too broad by user147263, callculus, TravisJ, SchrodingersCat, 6005 Nov 30 '15 at 18:44

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    $\begingroup$ How are you defining "accurate"? $\endgroup$ – Santiago Canez Nov 30 '15 at 17:21
  • $\begingroup$ There are standard formulae for estimating the remainder term. $\endgroup$ – Cheerful Parsnip Nov 30 '15 at 17:22
  • $\begingroup$ What's "accurate"? $\endgroup$ – 5xum Nov 30 '15 at 17:23
  • $\begingroup$ I suppose accuracy means closest to approximate. But below answer has already clarified. $\endgroup$ – chamburger Nov 30 '15 at 17:29
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It depends on the function, the desired accuracy and the interval in which you want the estimate to hold. Using the Lagrange form of the remainder, we have for the Taylor expansion on an interval $[a-\delta,a+\delta]$: $$ \Bigl|f(x)-\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}\,(x-a)^k\Bigl|\le\frac{\delta^{n+1}}{(n+1)!}\sup_{|x-a|\le\delta}|f^{(n+1)}(x)|,\quad|x-a|\le\delta. $$ The problem is usually to get a good estimate of $f^{(n+1)}$. This is easy for certain functions like $e^x$, $\cos x$ and $\sin x$.

In general, the smaller the desired accuracy and the larger the interval, the lager the $n$.

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A Taylor polynomial stops after the $x^n$ term, and then there is a remainder term that involves the $n+1^{th}$ derivative of the function.
You have to study that remainder term to decide how accurate the polynomial is.
The remainder is $$R_n=f^{(n+1)}(\xi)\frac{(x-x_o)^{n+1}}{(n+1)!}$$ for some number $\xi$ between $x$ and $x_o$.
For example, if $f(x)=\sin x$, then $|f^{(n+1)}(\xi)|<1$ because it is either a sine or a cosine, so $R_n<\frac{(x-x_o)^{n+1}}{(n+1)!}$ and this approaches zero as $n$ gets large, by the ratio test.

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  • $\begingroup$ Ah so its the remainder that determines the taylor series accuracy. Thanks. $\endgroup$ – chamburger Nov 30 '15 at 17:28

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