Π day

[Thomas Klemm]

(03-14-2022 03:35 AM)robve Wrote:  \( \frac{\pi}{6} = \frac{1}{\sqrt3}\left(1-\frac{1}{3^1\cdot3}+\frac{1}{3^2\cdot5}-\frac{1}{3^3\cdot7}+\cdots\right) \)

That's an alternating series so we can use Valentin's program:
(11C) Summation of infinite, alternating series

We write the program B to calculate the \(k\)-th term:

\(
\begin{align*}
a_k &= \frac{1}{3^k \, (2k + 1)}
\end{align*}
\)

Code:

084 - 42,21,12  LBL B
085 -        3  3
086 -       34  x<>y
087 -       14  y^x
088 -    43 36  LSTx
089 -        2  2
090 -       20  x
091 -        1  1
092 -       40  +
093 -       20  x
094 -       15  1/x
095 -    43 32  RTN

We use \(\text{PSum} = 10\), \(\text{NDif} = 7\) for maximum accuracy:

10 ENTER 7 A

After some time, we get the following result in the display:

0.9068996822

The correct result \(\frac{\pi}{\sqrt{12}}\) on the HP-11C is:

0.9068996821

---

Now you might be wondering: Isn't that like cracking a nut with a sledgehammer?

Let me use this program for the HP-42S instead:

Code:

00 { 17-Byte Prgm }
01▸LBL 00
02 1/X
03 LASTX
04 X<> ST Z
05 3
06 ÷
07 -
08 X<>Y
09 2
10 -
11 X>0?
12 GTO 00
13 R↓
14 END

We sum the terms backwards to improve accuracy.
It turns out that a starting value of \(43\) is good enough:

CLST
43
XEQ 00

We get:

0.906899682117

If we multiply that by \(\sqrt{12}\) we get:

3.14159265359

---

I must admit that I cheated a bit since I used this simulator for the HP-11C and Free42 instead of the real calculators.
Thus the results might be slightly off.