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# 9.2 Calculation of

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.

9.33   Lemma.

and

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.

9.34   Lemma (Limits of sine and cosine.) Let . Let be a sequence in such that . Then

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.

9.37   Theorem. Let be any sequence such that for all , and . Then
 (9.38)

Hence if for all we also have

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