Floating point number fractions excercise

Question

During my preparation for an exam, I've come across an exercise that was focused on precision calculation with floating point numbers. It goes like this:

Consider the expression : $$\frac{1}{1-x}-\frac{1}{1+x}, x\neq \pm1$$ For what range of values $x$ is it difficult to compute the expression accurately in foating-point arithmetic? Give a re-arrangement of the terms such that, for the range of values from part (a), the computation is more accurate in foating-point arithmetic.

I have some basics covered on floating point number systems, so I've tried to set $\pm x=0.9999...$, with a different number of decimal digits. The calculations are indeed more and more imprecise, however I don't know if this is the only problematic 'range'. Also, I have no idea how to transform this expression to be more accurate. I have tried $\frac{2x}{1-x^{2}}$ and $e^{\ln{2x}-\ln(1-x^{2})}$, but the results were pretty much the same.

Thanks for help!

Answer 1

Actually, floating point arithmetic according to the IEEE 754 standard will calculate 1-x exactly, without any rounding error, if 0.5 ≤ x ≤ 2.0. So for x close to 1, 1-x is calculated exactly, 1 / (1-x) is calculated accurately, and 1 / (1 + x) and the sum will be calculated accurately as well.

What you are likely observing is the rounding error when 0.999999 for example is converted to a floating point number. x will not be equal to 0.999999, but a bit off. And the function itself as a function of real numbers is very sensitive to changes in the argument. That also explains why re-arranging the function can't help, because the function is sensitive to the arguments, and the arguments are not what you think.

The same applies for x close to -1, since f(x) = f(-x).

You will actually have a problem if x is close to 0. 1 / (1-x) and 1 / (1+x) can both be calculated with a very small absolute error, the difference is calculated exactly because both have about the same size, but because the result is close to zero, the small absolute errors turn into a rather large relative error.

When x is close to zero, 1 / (1 - x) = 1 + x + x^2 + x^3 ..., and a / (1 + x) = 1 - x + x^2 - x^3 ..., the mathematical difference is 2x + 2x^3 + 2x^5 ... ≈ 2x.

Let u be the value of the last bit in a number 1 ≤ x < 2, then IEEE arithmetic has rounding errors up to u/2. 1 / (1 - u) ≥ 1 is calculated with an error up to u/2, 1 / (1 + u) has an error up to u/4. If you pick x^2 slightly larger than u/4, then 1 / (1 - x) is rounded to 1+x, 1 / (1 + x) is rounded to 1 - x + u/4, and the difference that is calculated is 2x - u/4, instead of 2x + x^3 + 2x^5 ..., so the absolute error is > u/4, and the relative error about $(u/4)^{1/2}$.

You can fix this easily by subtracting 1 from both sides, giving $x / (1 - x) + x / (1 + x)$.