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The proof of the next lemma depends on the following assumption,
which is explicitly
stated
by Archimedes [2, page 3]. This assumption
involves the ideas
of curve with given endpoints and length of curve (which I will leave
undiscussed).
9.32
Assumption.
Let
and
be points in
.
Then of all curves with endpoints
and
, the segment
is the shortest.
Proof:
Case 1: Suppose
.
Then (see the figure) the length of
the arc
joining to in the first and fourth quadrants is . (This
follows
from the definition of .) The length of the segment is .
By our assumption,
, i.e., . Since both and
are positive when
, we also have
.
Case 2: Suppose
. Then
so
and
in this case also. This proves
the first
assertion of lemma 9.33. If , then , so
Thus
and since the relation clearly holds when the lemma is proved.
Proof: By (9.25) we have
so
If , then
, so by the squeezing rule,
This means that
.
The proof that
is
similar.
The proof of the next lemma involves another new assumption.
9.35
Assumption.
Suppose
.
Let the tangent to the unit circle at
intersect the
axis at
, and let
.
Then the circular arc
joining
to
(and passing through
) is shorter than the curve made of
the two
segments
and
(see the figure).
Remark: Archimedes
makes a general assumption about curves that are concave
in the same direction [2, pages 2-4] which allows him to
prove our
assumption.
9.36
Lemma.
If
, then
Proof:
Suppose
. Draw the tangents to the unit circle at
and
and let the point at which they intersect the -axis be
. (By
symmetry
both tangents intersect the -axis at the same point.) Let
be the
point where
the segment intersects the -axis, and let
.
Triangles
and
are similar since they are right triangles with a
common acute
angle.
Hence
i.e.,
Now the length of the arc joining to is , and the length of
the
broken line from
to
to is
, so by
assumption 9.35,
i.e.,
This proves our lemma.
Proof: If
, then it follows from
lemma(9.36) that
. Since
it follows that
Hence by lemma 9.33 we have
|
(9.39) |
Let be a sequence for which for all
and . Then
we can find a number
such that for all
.
By (9.39)
By lemma
9.34, we know that
, so by the squeezing
rule
.
9.40
Example (Calculation of .)
Since
, it follows from
(
9.38) that
and hence that
This result can be used to find a good approximation to
. By the
half-angle
formula, we have
for
. Here I have used the fact that
for
. Also
so
By repeated applications of this process I can find
for arbitrary
, and then find
which will be a good approximation to
.
I wrote a set of Maple
routines to do
the calculations above. The procedure sinsq(n)
calculates
and the procedure
mypi(m)
calculates
. The `` fi" (which
is
`` if" spelled backwards) is Maple's way of ending an `` if" statement.
`` Digits " indicates that all
calculations are
done to decimal digits accuracy. The command `` evalf(Pi)" requests
the
decimal approximation to to be printed.
> sinsq :=
> n-> if n=1 then 1;
> else .5*(1-sqrt(1 - sinsq(n-1)));
> fi;
sinsq := proc(n) options operator,arrow; if n = 1 then 1
else .5 -.5*sqrt(1-sinsq(n-1)) fi end
> mypi := m -> 2^m*sqrt(sinsq(m));
> Digits := 20;
> mypi(4);
> mypi(8);
> mypi(12);
> mypi(16);
> mypi(20);
> mypi(24);
> mypi(28);
> mypi(32);
> mypi(36);
> mypi(40);
> evalf(Pi);
9.41
Exercise.
Examine the output of the program above. It appears that
. This
certainly
is not right. What can I conclude about
from my computer program?
9.42
Exercise.
Show that the number
is the area of
a regular
-gon inscribed in the unit circle. Make any reasonable geometric
assumptions, but
explain your ideas clearly.
Next: 9.3 Integrals of the
Up: 9. Trigonometric Functions
Previous: 9.1 Properties of Sine
  Index
Ray Mayer
2007-09-07