An ∞ n
2.
-
Nn v/s n
-
ln v/s n
-
Pn v/s n
-
An v/s n
3.
The trend seen in the above data and the graph is that with a successive increase in the stage, the number of sides increases by 4 times the previous number. The number of sides starts from 3, therefore,
a = 3 (1st term)
r = 4 (ratio between 2 successive terms)
Therefore, it is seen that the formula that generalizes the behaviour of the given set of values is:
ar n
Verification:
n = 0
arn
= 3 (4)0
= 3
n = 1
arn
= 3 (4)1
= 12
n = 2
arn
= 3 (4)3
= 48
Hence proved.
The trend seen in the above data and the graph is that with a successive increase in the stage, the length of 1 side decreases by 1/3 times the previous number. The length of 1 side starts at 1.
Therefore,
a = 1 (1st term)
r = 1/3 (ratio between 2 successive terms)
Therefore, it is seen that the formula that generalizes the behaviour of the given set of values is:
a r n
Verification:
n = 0
a r n
= 1 (1/3)0
= 1
n = 1
a r n
= 1 (1/3)1
= 1/3
n = 2
a r n
= 1 (1/3)2
= 1/9
Hence Proved.
The trend seen in the above data and the graph is that with each successive increase in the stage, the perimeter is equal to the number of sides × the length of 1 side; for that particular stage.
Therefore, it is seen that the formula that generalizes the behaviour of the given set of values is:
Pn = Nn × ln
Verification:
n = 0
Nn × ln
= 3 × 1
= 3
n = 1
Nn × ln
= 12 × 1/3
= 4
n = 2
Nn × ln
= 48 × 1/9
= 48/9
= 5 3/9
Hence proved.
The trend seen in the above data and the graph is that with each successive increase in the stage, the area increases by (3 × 4n-1)/9n.
Therefore, it is seen that the formula that generalizes the behaviour of the given set of values is:
An =
Where ‘n’ is the stage, k is the nth iteration of the Koch snowflake. The ‘s’ is taken as 1(i.e. from above), and is therefore not taken into consideration.
Verification:
n = 0
= /4
n = 1
=
=/3
n = 2
=
= 10/27
Hence Proved.
4. When n = 4;
-
Nn = a rn
N4 = 3 (4)4
= 768
-
ln = a rn
l4 = 1 (1/3)4
= 1/81
-
Pn = Nn × ln
Pn = 768 × 1
81
= 9.4815
-
An =
=
= 94/273
= 0.67001
One Side of the Fractal at Stage 4:
5. Calculation for values of N6, l6, P6 and A6 :
General Formula for Nn: a rn
a= 3
r= 4
n= 6
Therefore:
3 (4)6
= 12288
General Formula for ln: a rn
a= 1
r= 1/3
n= 6
Therefore:
1 (1/3)6
= 1/729
General formula for Pn: Pn = Nn × ln
Nn = 12288
ln = 1/729
Therefore:
12288× 1/729
=16.856
General Formula for An:
Therefore:
=1.3816
Therefore:
Successive terms of An in terms of A0 is:
P0 = 3 units
P1 = 4 units
P2 = units
P3 = units
P4 = 9.481481 units
A0 = unit²
A1 = unit²
A2 = unit²
A3 = unit²
A4 = unit2
Hence, as can be seen by the aforementioned values, as n gets larger, there is a geometric increase in the perimeter and area.
For perimeter, the geometric progression has a common ratio of 4/3. This is proved by:
The perimeter, as per the Perimeter graph, when n tends to infinity is also infinite.
For area, the geometric progression has a common ratio of.
So as:
A = A
=
=
=
=
= 0.692820323
Conclusion made about the area as:-
As n tends to, the area of the fractal geometrically progresses and the area becomes a constant, to six decimal places, when n = 17. Therefore we can conclude by saying that the snowflake has a finite area but an infinite perimeter.
Comment on the results obtained:
The results obtained while doing this seem to be accurate as there is no room for error in such an investigation. If any of the results obtained were not correct then there would have not been a common ratio for geometric progression in either of the cases, for the perimeter or the area. If there was no common ratio, there would have been no geometric progression, in which case, there would have been no relationship between the successive terms and the whole investigation would not make any sense.
Bibliography:
Sources from the net:
- www.ecademy.agnesscott.edu/
- www.google.com
Use of Technological Resources/ Software Packages:
- Graphic Display Calculator
- MathType
- Microsoft Word
- Microsoft Excel
In question 4, the drawing of 1 side of the 4th iteration of the Koch snowflake was drawn using Web Turtle:
http://www.sonic.net/~nbs/webturtle/webturtle.cgi
