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I know that the Taylor/Maclaurin(?) expansion for the sine function is

$$ \sin(x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dfrac{x^9}{9!} - \ ... $$

My questions are:

  1. If I have calculated the first, for example, 5 terms (the ones shown above) in the sine Taylor expansion, then at what values does the Taylor expansion have an error greater than 1% compared to the real sine function?

  2. How many terms would I need to calculate of the Taylor expansion for the sine function in order to have an error less than 1% at a certain point (e.g. $ x = 2\pi$)?

I asked this question elsewhere (not on this website) and was told to reasearch "Lagrange multipliers" but I couldn't understand it...can anyone here help me?

EDIT: No wonder I was lost, I was looking at the wrong theorem...

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    $\begingroup$ Not the "Lagrange-multipliers" is what you need, but the "Lagrange remainder term" $\endgroup$ – Peter Oct 9 '17 at 18:02
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Lagrange multipliers has nothing to do with this; the relevant term is the Lagrange remainder. This says that there exists a $\xi$ between $0$ and $x$ such that

$$f(x)-T_n(x)=\frac{f^{(n+1)}(\xi) x^{n+1}}{(n+1)!}$$ where $T_n$ is the Maclaurin approximant of $f$ of degree $n$. ("Maclaurin" just means that the point of Taylor expansion is $x=0$.)

In the case of sine specifically, that derivative can be uniformly bounded (regardless of what $\xi$, which we do not know, is) by $1$, so that

$$|f(x)-T_n(x)| \leq \frac{|x|^{n+1}}{(n+1)!}.$$

A relative error estimate like you ask for is problematic because $\sin(x)$ vanishes at certain points, so that your relative error estimate looks like

$$\frac{|f(x)-T_n(x)|}{|\sin(x)|} \leq \frac{|x|^{n+1}}{(n+1)! |\sin(x)|}$$

which is very awkwardly behaved near the zeroes of $\sin$ except for $0$. You can avoid the difficulty by restricting attention to the domain $[-\pi/2,\pi/2]$ and using periodicity to extract values elsewhere.

In the case of $\sin$ and also $\cos$, you can get a slightly better estimate than this one by using the alternating series error estimate, which says that if you have a sequence of positive numbers $a_n$ decreasing to zero, then $\left | \sum_{n=N}^\infty (-1)^n a_n \right | \leq a_N$. For $f=\sin$ you can use this to show that given an odd number $k$:

$$|f(x)-T_k(x)| \leq \frac{|x|^{k+2}}{(k+2)!}$$

provided $|x|<k+2$.

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  • $\begingroup$ Thanks, this helped me out. Now I know what to look for anyway. :) $\endgroup$ – Nick_2440 Oct 9 '17 at 18:36
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The power series for $\sin(x)$ and $\cos(x)$ for positive $x$ are enveloping. This means that the sum is between any two consecutive partial sums.

So, just compute partial sums until the last term added is less than your desired error and you are done.

I show how to prove this here:

prove that $ x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5 $ for some deleted neighborhood of $x=0$

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