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When n = 400, the value of π calculated from the area of the Inscribed polygon has 2 decimal place accuracy as compared to the given value of π, which is 3.141592654
As compared to the same given value of π, the value of π calculated from the area of the Circumscribed polygon has 3 decimal place accuracy when n = 400.
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Using the same method as used to calculate the above, we find that the Average Area has 4 decimal place accuracy
The decimal places of accuracy for both cases were found in the following way:
Table 1- Un-rounded Values:
Table 2- Rounded Values:
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.
The table above shows that the value of X is always lower that the value of Y, for all values of n, and therefore the value of Y – X is always positive.
Therefore, this proves that Area Y > Area X, and it is for this reason that the value of π calculated from the circumscribed polygon’s area converges faster towards the given (actual) value of π.
Estimation of π using half-perimeter method
Just as the area of a circle and two polygons, one inscribed, and the circumscribed, can be used to find the value of π, so can the perimeter.
The circumference of a circle is given by the formula C= 2πr (where C = circumference, r = radius). For a circle with a radius of 0.5cm, the circumference now becomes, C = 2×π×0.5 = π. So calculating the circumference of a circle with radius 0.5cm, would automatically give us the value of π. However, deriving and using a formula to find the circumference of a circle with radius 0.5 may become complicated. Using a radius of 1cm would be much simpler. If the circle has a radius of 1cm, its circumference will be 2π. However, if we calculate half the value of the circumference (semi-perimeter), we automatically calculate the value of π.
As before, finding the semi-perimeter of an inscribed and a circumscribed polygon with n-sides, and finding their average value will give an approximate value of the semi-perimeter of the circle, and therefore the value of π.
½Ci < π < ½Cc
where ½Ci = semi-perimeter of inscribed polygon and
½Cc = semi-perimeter of circumscribed polygon
Semi-Perimeter of Inscribed Polygon
The angle θ at the centre
As ,
In a n-sided polygon, there are n sides, and here a is half of one side,
Therefore the length of one side is 2a,
And the total perimeter of the polygon = 2an
Semi-perimeter of inscribed polygon
Semi-Perimeter of Circumscribed Polygon
The angle θ at the centre
As ,
In a n-sided polygon, there are n sides, and here a is half of one side,
Therefore the length of one side is 2a,
And the total perimeter of the polygon = 2an
Semi-perimeter of circumscribed polygon
Average Semi-Perimeter = Semi-Perimeter of Circumference ≈ π
and if
Average value of the semi-perimeters
Putting in values of n in a spreadsheet, the following results are obtained:
Graphs and Charts
Fig. 15
As can be seen from Figs. 15 and 16, which are graphs of n vs. the semi-perimeters, the graph of the inscribed polygon converges faster towards the actual value of π. This is opposite to the case of the π calculation by using areas of polygons, where the circumscribed polygon converged faster.
The reasoning for this is similar to the one before, but this time it applies to the inscribed polygon.
Degrees of Accuracy
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When n = 400, the value of π calculated from the semi-perimeter of the Inscribed polygon has 4 decimal place accuracy as compared to the given value of π, which is 3.141592654
As compared to the same given value of π, the value of π calculated from the semi-perimeter of the Circumscribed polygon has 3 decimal place accuracy when n = 400.
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Using the same method as used to calculate the above, we find that the Average Semi-Perimeter has 4 decimal place accuracy at n = 400.
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To get a calculated value of π to 6 decimal place accuracy:
A minimum value of n = 5801 was needed for the Inscribed polygon.
A minimum value of n = 3495 was needed for the Circumscribed polygon.
The above results were obtained in the same way as is mentioned in the previous section, of calculating π using the areas of polygons.
Comparing the degrees of accuracy of the value calculated by the circumscribed polygon, using the area and the semi-perimeter method, we can see that they are exactly the same. This is also true for the minimum value of 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.
