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# Compute π using the Taylor polynomial of arctan x

For this exercise define

1. Using the trig identity

twice, once with , and then the second time with , show that

Then use the same identity again with and to show

This establishes the identity

2. Using the Taylor polynomial approximation at prove that

3. Using the Taylor polynomial approximation at prove that

4. Using the above parts show that the value of to seven decimal places is 3.1415926.

1. Proof. Letting we have

Letting we have

Letting and we have (recalling that )

But then

2. Proof. We know the Taylor polynomial approximation for from this exercise (Section 7.8, Exercise #3):

Therefore, we can compute an approximation to ,

where

Therefore,

3. Proof. Again using the Taylor polynomial approximation to we have

4. Finally,

Prove

for all .

Proof. First,

And then,

# Use integrals to establish the formula for the area of an ellipse

Denoting the unit circle, , by we define an ellipse to be the set of points

1. Show that the points on satisfy the equation:

2. Prove that the area of is measurable and that

1. Proof. If is a point on then is a point on (since all points of are obtained by taking a point of and multiplying the -coordinate by and the -coordinate by ). By definition of , we must then have,

2. Proof. From part (a) we know is the set of points such that . But, this implies,

Hence, the area of is the area enclosed from to by the graphs of

To show this region is measurable and has area we begin with this identity (from Apostol, 2.4 Exercise 17),

In the second to last line we have used the expansion/contraction of the interval of integration. Hence, we know the integral from to of exists and has value . Thus, is measurable and

# Use dodecagons to deduce an inequality about π

By considering dodecagons inscribed and circumscribed about a unit disk, establish the inequalities

First, we draw some pictures of the situation for reference.

( Note: I don’t know a way to do this without using trig functions, which haven’t been introduced in the text yet. If you have an alternative approach without them, please leave a comment and let us know about it. )

Since these are dodecagons, the angle at the origin of the circle of each triangular sector is , and the angle of the right triangles formed by splitting each of these sectors in half (shown in the diagrams) is then . Then we use the fact that

Now, for the circumscribed dodecagon we have the area of the right triangle with base 1 in the diagram on the left given by

Since there are 24 such triangles in the dodecahedron, we then have the area of the circumscribed dodecahedron given by

For the inscribed dodecagon we consider the right triangle with hypotenuse 1 in the diagram. The length of one of the legs is then given by and the other is given by . So the area of the triangle is

Since there are 24 such triangles in the inscribed dodecahedron, we then have,

Since the area of the unit circle is, by definition, , and it lies in between these two dodecahedrons, we have,

# Compute some integrals in terms of π

Recall we have defined as the area of the unit disk. We have proved (Apostol, Section 2.3, Example 3) that

Use this and the theorems on the property of integrals to compute the following in terms of :

1. .
2. .
3. .

1. Using the expansion/contraction property of the integral (Apostol, Theorem 1.19) we compute:

2. Using the expansion/contraction property of the integeral (Apostol, Theorem 1.19) we compute:

3. Using the linearity properties of the integral, as well as, expansion/contraction and the formula for given above. We also use this exercise (Apostol, Section 1.25, Exercise #26 part (b)) which establishes that the integral from to of an odd function is 0.