An Inconsistency in Numerical Approximation

Question

Consider the expression

$$ 10^5 - \frac{10^{10}}{1+10^5}. $$

Using the elementary properties of fractions we can evaluate the expression as

$$ 10^5 - \frac{10^{10}}{1+10^5} = \frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = \frac{10^5}{1+10^5}\approx 1. $$

Note that the approximation $10^5+1 \approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get

$$ 10^5 - \frac{10^{10}}{1+10^5} \approx 10^5 - \frac{10^{10}}{10^5} = 0. $$

The same logic works for

$$ 10^p - \frac{10^{2p}}{1+10^p} $$

for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.

Is there an easy explanation of what's going on here?

Answer 1

It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is $$ x - \frac{x^2}{1+x}$$

Note that $$\frac{x^2}{1+x} = \frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$ so that $$ x - \frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$

In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.

Answer 2

The first approximation is fine. The second on is not, because, $10^5$ and $\dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10\,001$ is close to $10\,000$, then $1$ is close to $0$.

Answer 3

There is no paradox.

When you approximate $$\frac{10^5}{1+10^5}=1-0.000099999000\cdots$$ with $1$, the error is on the order of $10^{-5}$.

But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.