# Calculate $\sin(0.2)$ to 4 decimal places with its Taylor series and the Lagrange error bound

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.

## 3 Answers

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 $$n$$.

• Thank you! I had not realized that I had to use all the terms up to the nth one, hence the confusion when I was checking other examples on how to do it. – Lightsong Dec 30 '18 at 16:20

Since you found that $$n=3$$ is the degree of the Taylor polynomial that satisfies the error bound, your approximation should be $$\sin(0.2)=0.2-\frac{0.2^3}{3!},$$ in which you use $$P_3(x)$$ to approximate $$\sin(x)$$, based on your notation.

$$\sin x=x-\frac{x^3}{3!}+R_n$$ You calculated that for $$n=3$$ $$R_n<\varepsilon$$ where $$\varepsilon$$ is the error bound you have. Then $$n=3$$ is the degree of the Taylor polynomial that satisfies that error bound, so put $$x=0.2$$ into $$P_n$$ since that is what approximates the function, while $$R_n$$ tells you how good is the approximation of $$P_n$$.

You'll get $$\sin(0.2)=0.198669\ldots\simeq P_3=0.1986666\ldots$$