# Thread: MacLaurin series and AST error approximation

1. ## MacLaurin series and AST error approximation

I gotta show that

$\frac{1}{2!}-\frac{x^2}{4!} < \frac{1-cos(x)}{x^2} < \frac{1}{2!}$ for x not equal to 0

I know that $|S-S_{n}| < a_{n+1}$ and $S_{n-1} < S < S_{n}$ or $S_{n} < S < S{n-1}$ depending on which one is bigger value and smaller value

Using the MacLaurin series for cos x and applying $\frac{1-cos (x)}{x^2}$ to it, I got

$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!} = \frac{1}{2!} - \frac{x^2}{4!} + \frac{x^4}{6!}.....$

so clearly

$\frac{1}{2!} - \frac{x^2}{4!} < \,or\, > \frac{1 - cos(x)}{x^2} < \,or\, > \frac{1}{2!}$

what's the best way to show which of the two it is?

$\frac{1}{2!} - \frac{x^2}{4!} < \frac{1}{2!}$ for all values of x except x=0

^^ just use this argument?

2. ## Re: MacLaurin series and AST error approximation

The error in an alternating series that has been truncated to n terms is never any more than the absolute value of the next term.

So here if your series \displaystyle \begin{align*} \frac{1 - \cos{(x)}}{x^2} = \frac{1}{2!} - \frac{x^2}{4!} + \frac{x^4}{6!} - \dots \end{align*} is truncated to two terms, then \displaystyle \begin{align*} \frac{1 - \cos{(x)}}{x^2} \approx \frac{1}{2!} - \frac{x^2}{4!} \end{align*} with an error no greater than \displaystyle \begin{align*} \frac{x^4}{6!} \end{align*}.

So clearly \displaystyle \begin{align*} \frac{1}{2!} - \frac{x^2}{4!} < \frac{1 - \cos{(x)}}{x^2} \end{align*}, since the next term along is being added, thereby showing that your estimate is an UNDER estimate.

As for the other side of your inequality \displaystyle \begin{align*} \frac{1}{2!} - \frac{x^2}{4!} < \frac{1 - \cos{(x)}}{x^2} < \frac{1}{2!} \end{align*}, so where you are trying to show \displaystyle \begin{align*} \frac{1 - \cos{(x)}}{x^2} < \frac{1}{2!} \end{align*} it is EXACTLY the same thing, just with your series truncated to one term. What would be the error then? Would the truncated series be an under estimate or an over estimate?