Find an alternative expression for $y :=cos(x+\delta)-cos(x)$

Question

Explain the difficulty of computing $y :=cos(x+\delta)-cos(x)$ for small values of $\delta$. Find an alternative expression of $y$ that does not exhibit these difficulties.

So far I have figured out, as Delta is very small we get $cos(x+\delta)\approx cos(x)$ Thus we experience critical rounding errors (they would cancel each other out). I can't seem to find however a good alternative form that could get rid of this problem.

I hope someone can help me find one and maybe give me some explanation why this is a good alternative form.

Answer 1

The issues with computing $\cos (x+ \delta) - \cos x$ directly are the same as computing $(1+\delta) -1$. The computation of $\cos$ just adds a little more noise.

$\cos (x+ \delta) - \cos x = -2 \sin (x+ {\delta \over 2}) \sin {\delta \over 2}$.

This is exact in $\mathbb{R}$ arithmetic and there is no dogastrophic cancellation.

One way to think about it is to look at the relative error for small non zero $\delta$, with ideal computations but allowing for the fact that we have finite precision.

In particular, note that for small $\delta,\epsilon$ we have ${\cos (x+\delta+\epsilon)-\cos(x) \over \cos (x+\delta)-\cos(x) } -1 \approx {- (\sin x) (\delta+\epsilon) \over -(\sin x) (\delta)} -1 = {\epsilon \over \delta}$.

If $\epsilon$ and $\delta$ are of the same size, then we have a very large relative error.

Answer 2

From: $$\lim_{\delta \to 0} \frac{\cos(x+\delta)-\cos(x)}{\delta}=-\sin(x)$$ So: $$\frac{\cos(x+\delta)-\cos(x)}{\delta}\approx-\sin(x)$$ $$\cos(x+\delta)-\cos(x)\approx-\delta\sin(x)$$

Answer 3

Apply the difference-product relation for cosines:

$\cos u - \cos v = -2\sin (\frac{u+v}{2})\sin (\frac{u-v}{2})$

$\cos (x+\delta) - \cos x = -2\sin (\frac{2x+\delta}{2})\sin (\frac{\delta}{2})$