So, I have to estimate the value of sin(0,2) with at least 4 correct decimal places by using its Taylor series and the Lagrange error bound, but I am stuck in what I believe is a trivial step.
I understand the following:
$$\sin(x)=\sum_{n=0}^∞(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$f(x)=P_{n}(x) + R_{n}(x)$$ so, $$\sin(0,2)=P_{n}(0,2)+R_{n}(0,2)$$
By having that, I assume that I should find the value for n so that $$f^{(n+1}(x)\leq M$$ where M (I guess?) is 1, the maximum value for $\sin(x)$
Knowing that, My Lagrange error should be:
$$R_n(x)\leq\frac{M(x)^{n+1}}{(n+1)!}$$
so:
$$R_n(0,2)\leq\frac{(0,2)^{n+1}}{(n+1)!}\leq 0,0001$$
By checking values for n, I found that for n=3, its value is 0,00006…, and so I chose 3 as the degree of the Taylor polinomial that should give me an error smaller than 0,0001
Now I am stuck at calculating sin(0,2), which I assume it should be like this, where I do not know how to express Rn(x), nor if the n I chose is correct, since:
$$\sin(0,2)=\frac{(0,2)^{2n+1}}{(2n+1)!}=\frac{(0,2)^{(2*3)+1}}{((2*3)+1)!}=2.539682539682542*10^{-9}$$
which is nowhere near its "correct" value, which I found that it's best approximated by making n=0:
$$\sin(0,2)=\frac{(0,2)^{2n+1}}{(2n+1)!}=\frac{(0,2)}{1!}=0,2$$
If I take n=0, then I do not know how to even start writing Rn(x):
$$\sin(0,2)=\frac{(0,2)^{2n+1}}{(2n+1)!}=\frac{(0,2)}{1!}=>0,2-R_n(x)=0,2-what?$$
Any help and corrections are greatly appreciated, I am also unsure if I chose the value for n correctly.
Best Answer
It looks like you're using the 3rd-order term, but you need to use the whole series up to the 3rd-order term. Since the 0th- and 2nd- order terms are $0$, you just need to add the 1st-order term, which is $x$.
Second, to get 4 correct decimal places, the error really needs to be less than $0.00005.$
Third, if you're using $2n+1$ as the index in your series, be sure to use $2n+1$ in the remainder term, rather than just $n$.