The error in Taylor polynomial of $\sin x$ is $10^{-6}$...

Question

If we want to expand $f(x)=\sin x$ with Taylor polynomial centered at $x=0$. and error of that for$x=\frac{\pi}{2}$ be at most $10^{-6}$. then what order of polynomial should be at? ($n=?$)

As a hint our lecturer said we should use $c=\frac{\pi}{2}$ at formula of remainder of Taylor series.

Here is my work:

The reminder of Taylor series for the function $f(x)$ where $x_0=0$ is :

$$R_n(x)=\cfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$

And I estimate $\frac{\pi}{2}$ to closest integer which is $2$ for the value of $x$.

To find $n$ I should solve inequality:

$$|R_n(x)|\le 10^{-6}\quad\rightarrow\quad|\cfrac{f^{(n+1)}(\frac{\pi}{2})}{(n+1)!}\times2^{n+1}|\le 10^{-6}$$

$(n+1)$'st derivative of $f(x)=\sin x$ can be one of the $\pm\sin x$ ,$\pm\cos x$. So $|f^{(n+1)}(\frac{\pi}{2})|\in\{0,1\}$. If it is equal to zero then we have $0\le 10^{-6}$ and it is absolutely wrong so the best I can do is using $|f^{(n+1)}(\frac{\pi}{2})|=1$. hence the inequality is $\left|\cfrac{2^{n+1}}{(n+1)!}\right|\le 10^{-6}$ . After plugging in some $n$s I got $n\ge13$. so the order of the Taylor polynomial should be $13$.

Is my approach right ? as you can see in my answer I had some difficulties to find $|f^{(n+1)}(\frac{\pi}{2})|$ did I right at this point?

Answer 1

As $c$ can be anywhere in $[0,\frac\pi2]$, the best you can say is

$$\left|f^{(n+1)}(c)\right|\le1.$$

Now we can use the exact value of $\frac\pi2$ (unless a calculator is forbidden), and the possible remainders are bounded by the inverses of

$$8.11E-01, 3.94E+00, 4.79E+01, 1.09E+03, 3.97E+04, 2.12E+06 $$

and the order $11$ suffices.

Notice that as this is an alternating series, you need not precompute the number of terms, and you can stop when the current term becomes smaller than the requested error.

Answer 2

Write $$\sum_{n=0}^\infty\frac{(-1)^n }{(2 n+1)!}x^{2 n+1}=\sum_{n=0}^p\frac{(-1)^n }{(2 n+1)!}x^{2 n+1}+\sum_{n=p+1}^\infty\frac{(-1)^n }{(2 n+1)!}x^{2 n+1}$$ SInce it is an alternating series, you want to know $p$ such that $$R_p=\frac{x^{2 p+3}}{(2 p+3)!}\leq \epsilon\implies (2 p+3)! \geq\frac{x^{2 p+3}}{\epsilon}$$ If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.

Adapted to the present problem, this would give $$p\sim\frac{e \pi }{4}\,\exp\Big[W\left(-\frac{2 \log (\pi \epsilon )}{e \pi }\right)\Big]-\frac 74$$ where $W(.)$ is Lambert function. For sure, you need to use $\lceil p \rceil$.

For $\epsilon=10^{-6}$, this gives $p=4.31643$ and then $\lceil p \rceil=5$ (the exact solution would be $4.31727$). So, the expansion up to $n=11$ is sufficient, as @Yves Daoust already wrote (this would give $0.9999999437$ while one more term would lead to $1.000000001$).

Checking $$R_4=\frac{\pi ^{11}}{81749606400}\sim 3.60\times 10^{-6} > 10^{-6} $$ $$R_5=\frac{\pi ^{13}}{51011754393600}\sim 5.69\times 10^{-8} < 10^{-6} $$

If you cannot use Lambert function, since its argument is large, you could use the approximation given in the linked Wikipedia page $$W(x)= L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$. For your value, this would would give $W(.)\sim 1.07011$ while its exact value is $1.04434$.