Rewrite ${1 \over {1 + 2x}} - {{1 - x} \over {1+x}}$ to avoid loss of significance

Question

I would like to rewrite the term

$${1 \over {1 + 2x}} - {{1 - x} \over {1+x}}$$

in order to avoid a loss of significance for $x \approx 0$, meaning I would like to minimize the numerical error made when computing the fraction.

I think it might be wise to expand the left and the right term in order to write them as a single term. I receive:

$$2x^2 \over {1 + 3x + 2x^2}.$$

This doesn't look very good either, because $2x^2$ pushes $x \approx 0$ even further to $0$. I tried a polynomial division and received:

$$1 - {{3x + 1} \over {2x^2 + 3x + 1}}.$$

Now, there is at least the left term that is independent from $x$, but the right term might still cause problems. Is this already enough to avoid a loss of significance or do I need to try another step here?

The whole question is already mentioned above, it doesn't contain any further parts. — Julian, Apr 15 '17 at 13:45
I rephrased the question and used "loss of significance" instead. — Julian, Apr 15 '17 at 13:50
What do you mean by "loss of significance"? Are you looking to minimize the numerical error made when computing the fraction? — Piotr Benedysiuk, Apr 15 '17 at 13:53
Yes. :-) Sorry if the question is unclear, it's just poorly written by our instructor, I guess. — Julian, Apr 15 '17 at 13:54
Your second attempt loses significance: Computing for $x=1E-4$ with three digit mantissa, you obtain $0$! On the other hand, the first expression (also found by Piotr) produces the quite good $2E-8$ even with one digit mantissa. — Hagen von Eitzen, Apr 15 '17 at 14:08
I expanded my answer somewhat, although I am not sure on how correct it is. — Piotr Benedysiuk, Apr 15 '17 at 14:10

Piotr Benedysiuk · Accepted Answer · 2017-04-15T14:08:55.173

We have: $$\frac{1}{1+2x} - \frac{1-x}{1+x} = \frac{(1+x)}{(1+x)(1+2x)} - \frac{(1-x)(1+2x)}{(1+x)(1+2x)} = \frac{2x^2}{1 + 3x + 2x^2}$$ which indeed is the result you arrived at. This fraction is $0$ for $x = 0$ and close to zero we can say:

$$\frac{2x^2}{1 + 3x + 2x^2} \approx \mathcal{O}(x^2)$$

as $1 + 3x + 2x^2 \approx 1$ for $x \approx 0$.

Is this what you are looking for?

Edit: Seems to me this is the best way of approaching this problem. Let's say you want to compute the value at $x = 0.001$ and due to numerics you compute the value at $x = 0.001 + \epsilon$. Then the error will go like $\mathcal{O}(\epsilon^2)$ while we have

$\frac{1}{1+2x} = \mathcal{O}(x)$ and $\frac{1-x}{1+x} = \mathcal{O}(x)$ so you would make an error which is linear in $\epsilon$. Since $\epsilon << 1$ we have $\epsilon^2 < \epsilon$. Ofcourse this changes a bit if $\epsilon \approx x$ ie, when you want to compute values VERY close to zero. Then linear scaling in $\epsilon$ might actually be beneficial.

This makes sense, so my first attempt would be the whole answer already. — Julian, Apr 15 '17 at 14:23

score 0 · Answer 2 · answered Apr 15 '17 at 13:58

If you want to calculate $f(x)$ for really small values close to $0$, you can use derivatives. You can calculate the derivative of $f(x)$ and multiply it by your $x$. It is a pretty good approximation.

$\dfrac{df(0)}{dx}=\dfrac{-2}{(1+2x)^2}+\dfrac{2}{(1+x)^2}=0$

So since the first derivative is zero, you can calculate the second derivative and think of it as the acceleration in a position-time diagram:

$\dfrac{d^2 f(0)}{dx^2}=\dfrac{8}{(1+2x)^3}-\dfrac{4}{(1+x)^3}=4$

So we can approximate:($a$ is the acceleration or the second derivative of the diagram)

$f(x)\approx \dfrac{1}{2}ax^2\approx \dfrac{1}{2}*4*x^2\approx 2*x^2$

Thanks for your answer, but unfortunately, this is not what I am looking for. I don't want to change the term in order to approximate it, I just want to rewrite it, as I did above. — Julian, Apr 15 '17 at 14:04

score 0 · Answer 3 · answered Apr 15 '17 at 15:10

I also think that Taylor expansion is a good idea. In fact, we can write $${1 \over {1 + 2x}} - {{1 - x} \over {1+x}}=\frac{2x^2}{1 + 3x + 2x^2}=2x^2 \sum_{i=1}^\infty (-1)^{n-1} \left(2^n-1\right) x^{n-1}$$ and since this an alternating series, you will stop summing terms up to $p$ as soon as $$\left(2^{p+1}-1\right) x^p \leq \epsilon$$

Rewrite ${1 \over {1 + 2x}} - {{1 - x} \over {1+x}}$ to avoid loss of significance

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