Determination of 
The purpose of this problem is to get experience with MATLAB by writing a short section of code
to compute a sequence of approximations to  .
Model: We recognize  as the ratio of the circumference of a circle to its diameter. We can
approximate the circumference of the circle by the perimeter of regular polygons by a an
increasingly large number of sides. Since these perimeters converge to the circumference of the
circle, the ratios of the perimeters to the limiting circle's diameter should provide a sequence with
limit equal to  . Assume the circle has a radius of one (i.e., a "unit circle") and thus a diameter of 2.
Let pn represent the perimeter of a regular polygon with 2 n sides that is inscribed in the unit circle.
We claim:
lim
n
n=2
pn

2
n=3
n=4
Analysis: We know that if we can determine the length of a side of the regular polygon of 2 n sides
that is inscribed in the unit circle (call this length sn ) then pn  2 n sn . Simple geometry for the
inscribed square tells us that s2  2 . Our idea is to use the value of s2 to determine s3 , the side
of the inscribed regular octagon, then to use that to get s4 , the side of the inscribed regular
hexadecagon, and so on. The general step is to use sn to obtain sn1 .
Consider the following depiction of the inscribed polygon with
The side of the polygon of 2 n1 sides is indicated and is obtained
bisecting the arc subtended by the side of length sn .:
Noticing the two right triangles and applying the Pythagorean
Theorem twice, we have:
2
sn
2
(1  x ) 
1
2
and
2
s
x 2  n  sn21
2
F
IJ
G
HK
F
I
G
HJ
K
2 n sides.
by
Solving for x in the first equation, we have
Fs I
x  1  1  GJ
H2 K
s II
Fs I F1  F
 1  2 1  GJ  G
G
H2 K H H2 J
KJ
K
Fs I Fs I
 2  2 1  GJ  GJ
H2 K H2 K
2
n
Thus
2
x
2
2
n
n
2
2
n
n
and by substituting this into the second equation above, we finally obtain:
s I
F
G
H2 J
K s
2
2  2 1
2
n 1
n
or
F
G
G
H
s I I
F
G
J
H2 J
KJ
K
2
sn 1  2 1  1 
n
Remembering that pn  2 n sn and thus pn 1  2n 1 sn1 , we can remove the variables representing the
sides and express everything in terms of the polygon perimeters:
F
G
G
H
F
IJIJ
G
H KJ
K
F
G
G
H
I
F
I
G
J
H J
KJ
K
pn 1
pn
 2 1 1
n 1
2
2  2n
that is
2
2
pn 1
pn
 2n 2 1  1 
2
2  2n
pn
for n  2,3,... is the sequence of quantities that converges to  , let is use the
2
label Pn for this approximation to  , and finally we obtain the recurrence relation that
and recalling that
Pn 1  2
n
F
2G
1
G
H
FP I I
1  GJJ
H2 KJ
K
2
n
n
for n  2,3,...
p2 4 s2

 2 2 , which is the approximation to  based upon the
which we start with P2 
2
2
inscribed square.
Problem: Write a short section of Matlab code that uses the recurrence above to produce
P2 , P3 ,..., P40 . Make sure you use the "format long e" command to see all of the digits.