# Thread: Lagrange error bound to estimate sin4° to five decimal places( maclaurin series)

1. ## Lagrange error bound to estimate sin4° to five decimal places( maclaurin series)

Lagrange error bound to estimate sin4° to five decimal places( maclaurin series)
|Rn(pi/45)<1*(pi/45)^n+1/(n+1)! < 5*10^-6
and the answer key says n should be greater than or equal to 3.

It doesn't make sense .

Because, if you write out derivatives, the ones with sines will disappear in the polynomial. So, don't we have to ignore sin (since it is maclaurin series)
So if it is 7rd order polynomial it should be x-(1/3!)x^3 + (x^5)/5!) -(x^7)/7!.
and therefore we need to look at 9th derivative.

It seems the answer key just applied the remainder theorem.

2. Originally Posted by hangainlover
Lagrange error bound to estimate sin4° to five decimal places( maclaurin series)
|Rn(pi/45)<1*(pi/45)^n+1/(n+1)! < 5*10^-6
and the answer key says n should be greater than or equal to 3.

It doesn't make sense .

Because, if you write out derivatives, the ones with sines will disappear in the polynomial. So, don't we have to ignore sin (since it is maclaurin series)
So if it is 7rd order polynomial it should be x-(1/3!)x^3 + (x^5)/5!) -(x^7)/7!.
and therefore we need to look at 9th derivative.

It seems the answer key just applied the remainder theorem.
$|R_n(x)|= \frac{\max_{u \in [0,x]}(|f^{(n+1)}(u)| |x^{n+1}|}{(n+1)!}$

As $f(x)=\sin(x)$ we may use $|f^{(k)}| \le 1$ to get:

$|R_n(x)|\le \frac{ |x^{n+1}|}{(n+1)!}$.

So now find $n$ such that:

$\frac{ |(\pi/45)^{n+1}|}{(n+1)!}<5*10^{-6}$.

(you do this last part by checking a few values of $n$)

If you don't beleive the book answer compare $\sin(\pi/45)$ with $\pi/45$ and $(\pi/45)-\frac{(\pi/45)^3}{3!}$

CB