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• Level: GCSE
• Subject: Maths
• Word count: 2482

# Estimation of &amp;eth;

Extracts from this document...

Introduction

Shayan Ghosh

IB HL Mathematics

3rd May 2002

## Estimation of π

Finding an approximate value of π by finding the area of a circle with radius of 1cm. This estimation will be done by calculating the areas of two n-sided polygons, one inscribed and the other circumscribed.

Fig. 1

1. The area of a circle is given by the formula A= πr² (where A = area, r = radius). For a circle with a radius of 1cm, the area now becomes, A = π × 1² = π.  So if we knew the area of a circle with a radius of 1cm, we would automatically get the value of π.
The sides of the circumscribed polygon touch the circle at parts just slightly.  However, as the sides of the polygon touch the circle from the outside, the area of the circumscribed polygon is slightly greater than that of the circle.
The inscribed polygon’s vertices touch the circle from inside; therefore the area of the inscribed polygon is slightly less than that of the circle.

Mathematically therefore, A
i < π < Ac          Ai = area of inscribed polygon and
A
c = area of circumscribed polygon

As the number of sides of the polygon,
n, grows, the difference in areas between each polygon and the circle will decrease, and therefore the average value of the areas will converge onto the actual value of π.
2.

Fig. 2

Middle

0

1

2

3

4

5

6

7

8

Given Value

3

.

1

4

1

5

9

2

6

5

Inscribed

3

.

1

4

1

4

6

3

4

6

Circumscribed

3

.

1

4

1

6

5

7

2

5

Average

3

.

1

4

1

5

6

0

3

6

Table 2- Rounded Values:

 Decimal Place 0 1 2 3 Given Value 3 . 1 4 2 Inscribed 3 . 1 4 1 Circumscribed 3 . 1 4 2
 Given Value 3 . 1 4 1 6 Average 3 . 1 4 1 6

The tables above shows (1) the two calculated values of π at n = 400 against the given value, and (2) the digit for each place value. Table 1 gives the three numbers just as were calculated or given, without any modifications. Table 2 shows each number rounded off to three decimal places.

To find the decimal places of accuracy for each calculated value of π, look at Table 2 and simply compare how many digits after the decimal point of the calculated value are the same as that of the given value, and read off the number from the row Decimal Places.

On the spreadsheet, each cell containing a calculated or entered value was set to show only 6 decimal places (in Microsoft Excel, the  buttons were used).

After having done this, the values of n were extended in the spreadsheet till the first 6 decimal place accurate number, as compared to the given value, is reached. This was done separately for each polygon.

To get a calculated value of π
to 6 decimal place accuracy:

A minimum value of n = 11602 was needed for the Inscribed polygon.
A minimum value of
n = 3495 was needed for the Circumscribed polygon.

Graphs and Charts

Looking at Figures 6 and 7, which are the graphs of the n vs. Areas, and n vs. Difference in Calculated and Actual Values of π respectively, we can see that the area of the circumscribed polygon converges faster towards the actual value of π. The reason for this can be seen by drawing a scale drawing of Fig. 1.

As the figure shows,

Area Y > Area X

This can be shown quantitatively too. We shall first derive expressions for each shaded area (X and Y) separately.

Area Y is the area of a segment (ABC) of the circle.

Area Y         = Area of Sector ABCD – Area of ∆ACD

where

and as r = 1

For Area X it is slightly more complicated.

Area X         = (Area Y + Area X) – Area Y

There are n (X + Y)s in a polygon of n sides (see Fig.10).

Therefore,

And as r = 1,

And as 180° = 3.14 in radians,

Now that we have expressions for both X and Y, entering the formulae into a spreadsheet for different values of n, we should be able to validate the inequality.

 n Y X Difference (Y-X) 5 0.1525702697 0.0980559690 0.0545143008 25 0.0013167621 0.0006646561 0.0006521060 50 0.0001649852 0.0000826882 0.0000822971 75 0.0000489059 0.0000244787 0.0000244272 100 0.0000206354 0.0000103238 0.0000103116 125 0.0000105661 0.0000052850 0.0000052810 150 0.0000061149 0.0000030582 0.0000030566 175 0.0000038508 0.0000019258 0.0000019250 200 0.0000025798 0.0000012901 0.0000012897 225 0.0000018119 0.0000009061 0.0000009058 250 0.0000013209 0.0000006605 0.0000006604 275 0.0000009924 0.0000004962 0.0000004962 300 0.0000007644 0.0000003822 0.0000003822 325 0.0000006012 0.0000003006 0.0000003006 350 0.0000004814 0.0000002407 0.0000002407 375 0.0000003914 0.0000001957 0.0000001957 400 0.0000003225 0.0000001612 0.0000001612

Conclusion

n required for six decimal place accuracy.

The reason for this is quite simple. When r = 1, the two formula become the same. Due to this all calculations therefore also produce equal results.

The surprising thing however is that when dealing with the inscribed polygon, the degree of accuracy doubles when using the semi-perimeter method.  This is further demonstrated by the minimum value of n required for 6 decimal place accuracy. Using the area method, the minimum value of n was over 11600, whereas using the semi-perimeter method, this minimum value is halved to 5800.

This increase in accuracy must be related to the difference in its formulae.

Area of inscribed polygon

Semi-perimeter of inscribed polygon

When r = 1, the formulae reduce, and the area formula has a constant of 0.5 being multiplied, and a, whereas the semi-perimeter formula has a constant of 1 being multiplied to it, and a .

But more importantly, the semi-perimeter formula has a  term in it, whereas the area formula has a term.  As 360 is double 180, and the same number, n, is dividing both 360 and 180, the formula with the term converges twice as fast towards its final value, i.e. π.

The combined effect of these two differences in the formulae results in the difference of the rate of convergence towards π of the values produced by each, and thus in their accuracy levels.

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