Originally Posted by

**Glitch** **The question:**

Find a value for n such that the absolute error in the approximation $\displaystyle f(2) \approx p_n(2)$ is less than $\displaystyle 10^{-6}$, given that $\displaystyle |R_{n+1}(x)| = |\frac{sin(c) x^{n + 1}}{(n + 1)!}|$ about 0.

(Note: f(x) = cos(x))

**My attempt:**

I replace sin(c) with 1, since it is an upper bound, and x with 2:

$\displaystyle |\frac{2^{n + 1}}{(n + 1)!}|$

We want this error to be less that $\displaystyle 10^{-6}$, so:

$\displaystyle |\frac{2^{n + 1}}{(n + 1)!}| < 10^{-6}$

This is equivalent to finding:

$\displaystyle 2^{n + 1} < 10^{-6}$

Why?! Of course that then you get what you get since the LHS is the power of a number greater than 2, whereas the RHS

is a **negative** power of a number greater than 1...! You forgot, or for some reason you

thought, that you can dismiss the LHS's denominator:

$\displaystyle \displaystyle{\frac{2^{n+1}}{(n+1)!}<10^{-6}\Longrightarrow (n+1)\ln 2<-6\ln 10+\sum\limits^{n+1}_{k=1}\ln k<-6\ln 10+(n+1)\ln (n+1)\Longrightarrow}$

$\displaystyle \displaystyle{\Longrightarrow (n+1)[\ln 2-\ln(n+1]<-6\ln 10\Longrightarrow n+1>\frac{6\ln 10}{\ln 2-\ln(n+1)}>\frac{6\ln 10}{\ln 2} \cong 19.9\Longrightarrow}$

$\displaystyle \displaystyle{ n>18.9}$

So I get $\displaystyle n=19$ .

Tonio

$\displaystyle (n + 1)ln(2) < -6ln(10)$

$\displaystyle (n + 1) < \frac{-6ln(10)}{ln(2)}$

$\displaystyle n < -20.93$

This isn't right. The answer should be n = 14. Where have I gone wrong? Thanks.