# Poorly conditioned, easily evaluated sum for unit testing

I am looking for examples of poorly conditioned sums which can rapidly be evaluated, for the purposes of unit testing.

I'm currently using the series representation for $$\ln(2)$$:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$$ which has an infinite summation condition number, but in double precision, I can only get the condition number to ~22 after an obscene number of terms (terrible for a unit test), and the error in the naive summation is not large. So I guess that this sum is numerically well-conditioned!

• Could you explain what you mean by the condition number of the sum?
– smh
Jan 29, 2019 at 14:48
• @smh I have included a few words on the conditioning of sums in my answer. The key is that if $s = a+b \approx 0$, then relatively small changes to $a$ and $b$ can lead to a relative large change in $s$. Example: If a company is making only a small profit, then a small increase in their income can cause their profits to, say, triple. Jan 30, 2019 at 14:55

The condition number of sum $$s(x) = \sum_{j=1}^n x_j$$ is given by $$\kappa(x) = \frac{\sum_{j=1}^n |x_j|}{|\sum_{j=1}^n x_j|} = \frac{s(|x|)}{|s(x)|}$$ and reflects the sums sensitivity to small changes in the input. Specifically, we have $$\underset{\epsilon \rightarrow 0_+}{\lim}\sup \left\{ \frac{1}{\epsilon} \left|\frac{s(x+\Delta x) - s(x)}{s(x)} \right| \: : \: |\Delta x_j| \leq \epsilon |x_j|\right\} = \kappa(x).$$ If all $$x_j$$ have the same sign, then the sum is well-conditioned. If the $$s(x) \approx 0$$, but $$s(|x|)$$ is large, then the sum is ill-conditioned. A good example of a sum whose conditioning is well understood is the polynomial $$p_n$$ given by $$p_n(x) = \sum_{j=0}^n \frac{x^j}{j!}$$ This is the Taylor polynomial of order $$n$$ for the natural exponential function at the point $$x_0 = 0$$. The sum is ill-conditioned for large negative values of $$x$$ and well-conditioned for all positive values of $$x$$. If $$x$$ is positive and $$n$$ is so large that $$p_n(x) \approx e^x$$ is a good approximation, then $$1/e^{x} \approx p_n(-x)$$ is a good approximation as well. In short, you have an ill-conditioned sum for which a good approximation is easy to obtain.
• @user14717 You are very welcome. I have fixed the index error in the definition of $p_n$. Jan 30, 2019 at 14:58