(b) What is the area enclosed by the
n
th curve in this series? (Note that the
n
th curve
in this series is obtained from the (
n

1)st curve by adding a large number of small squares.
For example, we obtain the second curve from the first one by adding four squares, each of
which has side length 1
/
3 and thus area 1
/
9. So the area enclosed by the second curve is
1 + 4
/
9.) Your answer should involve summing a geometric series.
Solution.
To obtain the
n
th curve from the (
n

1)st curve, note that the (
n

1)st curve
consists of 4
·
5
n

2
line segments of length (1
/
3)
n

2
. On each of these line segments we will
place a square of side length (1
/
3)
n

1
, and thus of area (1
/
9)
n

1
. So to get from the (
n

1)st
curve to the
n
th curve, we add squares having total area 4
·
5
n

2
·
(1
/
9)
n

1
; we can rearrange
this to get (4
/
5)
·
(5
/
9)
n

1
.
The area enclosed by the
n
th curve in this series (for
n
≥
2) is therefore
1 + (4
/
5)(5
/
9) + (4
/
5)(5
/
9)
2
+
· · ·
+ (4
/
5)(5
/
9)
n

1
and all the terms but the first make up a geometric series. We can rewrite this as
1 +
h
(4
/
9) + (4
/
9)(5
/
9) + (4
/
9)(5
/
9)
2
+
· · ·
+ (4
/
9)(5
/
9)
n

2
i
and recall that
a
+
ar
+
. . .
+
ar
k
=
a
(1

r
k
+1
)
/
(1

r
). With
a
= 4
/
9
, r
= 5
/
9
, k
=
n

2,
the area enclosed is therefore
1 + 4
/
9
1

(5
/
9)
n

1
1

5
/
9
!
This can be simplified to give
1 + (1

(5
/
9)
n

1
)
and if
n
is very large this is very nearly 2. In fact, the area enclosed by this curve is very
close to being a square of side length
√
2.
6.
Start with the point 0.
Flip a coin.
If it comes up heads, move 2
/
3 of the way
toward 1, if it comes up tails, move 2
/
3 of the way to 0 from wherever you are at the
time. Repeat forever. The points you find are drawing a picture of the Cantor set. Verify
that any point in the Cantor set will move to another point in the Cantor set under the
coinflippingandmoving process. (B+S 6.3.26)
Solution.
Moving twothirds of the way towards 1 takes the point
x
to (2 +
x
)
/
3. To see
this, note that the distance from
x
to 1 is 1

x
; the distance from (2 +
x
)
/
3 to 1 is
1

2 +
x
3
=
1

x
3
.
4
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Moving twothirds of the way towards 0 takes the point
x
to
x/
3.
So we need to show that if
x
is in the Cantor set, then so are (2 +
x
)
/
3 and
x/
3. But we
did this in class, by looking at the ternary expansion of
x
.
5
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '09
 Lugo
 Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

Click to edit the document details