I need help. I have $f(x)=sin(x)$. If I want to use Lagrange polynomial to make an approximation of $f(x)$, what should be the degree of that polynomial if I work in the interval $[0,\pi]$, and the error, $|sin(x)-L_n(x)|$ must be lesser or equal than 0.001 ?
In other words:$$|sin(x)-L_n(x)|={sin^{(n+1)}(c)\over(n+1)!}(x-x_o)...(x-x_n)\le0.001\ with \ 0\le c \le \pi $$
How to calculate $n$.
On
Based on your question providing no points then gammatesters answer seems like the answer you want but assuming we picked some equidistant points and intervals of 1 / 2 such as
$$x_i = 0, \frac{1}{2},1,\frac{3}{2},2...3 $$
we could fit the polynomial
$f(x)=$
$a+(x-3)(b+(c+(d+(e+(f-g(x - h))(x-j))(x - .5))(x-k))x)$
with
$ a=.141120008$
$ b=.047040002$
$ c=-.411971103$
$d=-.0472466735$
$e=.037260033506$
$f=.00119275707$
$g=.001301301$
$h=1$
$j=2.5$
$k=1.5$
The residuals are much less than .001 as the graph of them shows
.
On
Ok, I've thinking in the question and I've conclude that:
$$\left|{sin^{(n+1)}(c)\over(n+1)!}\right| \le 0.001 = {\left|sin^{(n+1)}(c)\right|\over(n+1)!}\le 0.001$$
Given that the range of $f(x), f'(x),f''(x),...,f^{(n+1)}(x)$ is $-1 \le y \le 1$ and the working interval is $[0,\pi]$, thus $\left|sin^{(n+1)}(c)\right| \le 1 $. Then ${1\over{(n+1)!}}\le0.001 $.
Solving this inequation we can conclude that $(n+1)! \ge 1000$, therefore, $n=6$
I think I'm right but I'm not 100% sure.
Computation of the mininum $n$ seems difficult, especially because you give no information about the $x_i.$ If they are pair-wise different, you are on the safe side if you assume $$\left| \sin(x)-L_n(x)\right| \le\left| \frac{\sin^{(n+1)}(c)}{(n+1)!}(x-x_o)...(x-x_n)\right| \le \frac{\pi^{n+1}}{(n+1)!} =: E_n$$ and then compute the $E_n, n=1,2,\dots$ until $E_n< 0.01\;$ (or is it $0.001?$)
In any case $n=12\;$ should work, because $E_{10}\approx 0.00737,\;$ $E_{11}\approx 0.00192957,\;$ and $E_{12}\approx 0.0004663.$