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Meik Hellmund
2026-03-05 20:09:16 +01:00
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chapters/Pi2.qmd
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@@ -16,13 +16,15 @@ using InteractiveUtils
flush(stdout)
```
# A Case Study in Floating-Point Arithmetic Stability
# A Case Study on Floating-Point Arithmetic Stability
This chapter is inspired by the example in *Christoph Überhuber*, "Computer-Numerik" Vol. 1, Springer 1995, Chap. 2.3.
## Calculation of $\pi$ According to Archimedes
A lower bound for $2\pi$—the circumference of the unit circle—is obtained by summing
A lower bound for $2\pi$ -- the circumference of the unit circle -- is obtained by summing
the side lengths of a regular $n$-gon inscribed in the unit circle.
The figure on the left shows how one can iteratively double the number of vertices starting from a square with side length $$s_4=\sqrt{2}$$.
The figure on the left illustrates the iterative doubling of the number of vertices starting from a square with side length $$s_4=\sqrt{2}$$.
:::{.narrow}
@@ -39,7 +41,7 @@ With
$|\overline{AC}|= s_{n},\quad |\overline{AB}|= |\overline{BC}|= s_{2n},\quad |\overline{MN}| =a, \quad |\overline{NB}| =1-a,$ the Pythagorean theorem applied to triangles $MNA$ and
$NBA$ respectively yields
$$
a^2 + \left(\frac{s_{n}}{2}\right)^2 = 1\qquad\text{resp.} \qquad
a^2 + \left(\frac{s_{n}}{2}\right)^2 = 1\qquad\text{and } \qquad
(1-a)^2 + \left(\frac{s_{n}}{2}\right)^2 = s_{2n}^2
$$
Elimination of $a$ gives the recursion
@@ -47,8 +49,7 @@ $$
s_{2n} = \sqrt{2-\sqrt{4-s_n^2}} \qquad\text{with initial value}\qquad
s_4=\sqrt{2}
$$
for the length $s_n$ of one side of the inscribed regular
$n$-gon.
for the length $s_n$ of one side of the inscribed regular $n$-gon.
The sequence $(n\cdot s_n)$ converges monotonically from below to the limit $2\pi$:
$$
@@ -59,7 +60,9 @@ $$
\epsilon_n = \left| \frac{n\, s_n-2 \pi}{2\pi} \right|
$$
## Two Iteration Formulas^[by Christoph Überhuber, "Computer-Numerik" Vol. 1, Springer 1995, Chap. 2.3]. The equation
## Two Iteration Formulas
The equation
$$
s_{2n} = \sqrt{2-\sqrt{4-s_n^2}}\qquad \qquad \text{(Iteration A)}
$$
@@ -68,9 +71,9 @@ $$
s_{2n} = \frac{s_n}{\sqrt{2+\sqrt{4-s_n^2}}} \qquad \qquad \text{(Iteration B)}
$$
(Please verify!)
However, Iteration A is ill-conditioned and numerically unstable, as the following code demonstrates. The output shows the respective approximation for $\pi$.
However, Iteration A is ill-conditioned and numerically unstable, as the following code demonstrates. The output shows the approximation for $\pi$ for both formulae iteration.
```{julia}
using Printf
@@ -102,7 +105,7 @@ for i in 3:35
end
```
While Iteration B stabilizes at a value correct within machine precision, Iteration A becomes unstable. A plot of the relative errors $\epsilon_i$ confirms this:
While Iteration B stabilizes to a value accurate within machine precision, Iteration A is unstable and diverges. A plot of the relative errors $\epsilon_i$ confirms this:
```{julia}
using PlotlyJS
@@ -132,8 +135,8 @@ Further iterations do not improve the result; it stabilizes at a relative error
Iteration A is unstable; already at $i=16$, the relative error begins to grow again.
The cause is a typical cancellation effect. The side lengths $s_n$ become very small, very quickly. Thus
$a_n=\sqrt{4-s_n^2}$ is only slightly smaller than 2, and when computing $s_{2n}=\sqrt{2-a_n}$, a typical cancellation effect occurs.
The cause is a typical numerical cancellation effect. The side lengths $s_n$ become very small very quickly. Thus
$a_n=\sqrt{4-s_n^2}$ is only slightly smaller than 2, and computing $s_{2n}=\sqrt{2-a_n}$ leads to a catastrophic cancellation.
```{julia}
setprecision(80) # precision for BigFloat
@@ -142,7 +145,7 @@ s = sqrt(BigFloat(2))
@printf " a = √(4-s^2) as BigFloat and as Float64\n\n"
for $$i=$ 3:44$
for i= 3:44
s = iterationA(s)
x = sqrt(4-s^2)
if i > 20
@@ -151,7 +154,8 @@ for $$i=$ 3:44$
end
```
One sees the decrease in the number of significant digits. One also sees that using `BigFloat` with a mantissa length of 80 bits only slightly delays the onset of the cancellation effect.
This demonstrates the loss of significant digits. It also shows that using `BigFloat` with 80 bits of precission (mantissa length) only slightly delays
the onset of the cancellation effect.
**Countermeasures against unstable algorithms typically require better, stable algorithms, not more precise machine numbers.**