- Level: International Baccalaureate
- Subject: Maths
- Word count: 1643
Math portfolio stellar numbers. This assessment will investigate geometric shapes that lead to special numbers.
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Introduction
Math Portfolio
Type 1
Stellar numbers
Student: Alehsan petersen
Class: IB08A
This assessment will investigate geometric shapes that lead to special numbers.
An example of this would be square numbers (1,4,9,16…), which can be represented by squares of side (1,2,3,4…).
First, triangles will be investigated.
The number of dots (1,3,6,10,15…) in each triangle is a triangular number, which will be labelled Tn, the triangular sequence of n. One can see that these triangles are equilateral triangles, where the nth triangle has n number of dots on each side.
It can be observed that the increase in dots follows the pattern1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+…+n. Therefore one can derive other numbers for Tn.
n | Tn | difference d | difference d2 |
1 | 1 | ||
2 | 3 | 2 | |
3 | 6 | 3 | 1 |
4 | 10 | 4 | 1 |
5 | 15 | 5 | 1 |
6 | 20 | 6 | 1 |
7 | 27 | 7 | 1 |
8 | 36 | 8 | 1 |
9 | 45 | 9 | 1 |
10 | 55 | 10 | 1 |
Since one is dealing with a triangular shape it would be likely to investigate if the double amount of dots, a rectangular shape, gives a pattern. The sequence of the rectangle is called Rn.( 2,6,12,20,30……)
Observing Rn and the table one realizes that ever since d2 seems to be an arithmetic sequence, it is indicated that Tn itself could have a quadratic formula as its generator.Actually,one can that the square of n plus n gives Rn. Expressing this mathematically,
Rn =
Since Rn=2Tn one can substitute Rn in order to get Tn,
2Tn=
2Tn=n(n+1)
Tn=
Middle
The method that was used for the table heading in cellA1 and B1. And made an excel formula in A2 namely A1-B1. So the difference was obtained. For the graph A1 was plotted in relation to B1.
One can observe from the graph and the table that since difference of Sn values seem to be an arithmetic sequence then Sn is likely to be generated by a quadratic formula. Whith this knowledge following process is obtained.
Observing Sn one realizes that every value of Sn can be expressed as the difference of the difference times a number plus 1. So,
Sn=
From the table it can be assumed that is 12 consistently; a value for can be inserted,
Sn=12
Now x is investigated, one observes that the values of x are of some kind of number sequence which will be called Cn. Cn will be investigated by exemplifying with the value of S7.
S7 = 253
S7= 252+1
S7= 12 21 +1
In order to receive a general statement this expression will be wrote in terms of n.
Sn = 12 Cn+1
Calculating the values for Cn following values are obtained.
(0,1,3,6,10,15…)
Looking at this sequence one notices that it seems to be similar to the triangular sequence derived previously. In fact, observing and comparing the sequences one can see that the Cn corresponds to T(n-1).
Now I can substitute Cn with T(n-1)
Thus,
Sn= 12 T(n-1)+1
Since,
Tn =
Sn will be,
Looking at the general statement of the star it can be noticed that the coefficient 6 is equal to the number of vertices in the stellar shape. Assuming 6 represents p, a general statement in terms of p and n that generates the sequence of p-stellar numbers for any value of p at stage Sn can be derived. Expressing this mathematically,
Sn=pn(n-1)+1
Testing this hypothesis requires an investigation of the vertices-value p. First, a star with five vertices is investigated.
When p =5 the star has a five stellar number with the following values for Sn.
S1=1
S2=11
S3=31
S4=61
stellar n | nr of dots | difference d | difference d2 |
1 | 1 | ||
2 | 11 | 10 | |
3 | 31 | 20 | 10 |
4 | 61 | 30 | 10 |
5 | 101 | 40 | 10 |
6 | 151 | 50 | 10 |
7 | 211 | 60 | 10 |
8 | 281 | 70 | 10 |
9 | 361 | 80 | 10 |
10 | 451 | 90 | 10 |
Conclusion
Furthermore, investigating negative values, one can observe that for the general statement Sn=pn(n-1)+1
If p=negativeSn=negative
If n=negativeSn=negative
If p=negative and n=negativeSn=positive
It seems that theoretically any rational number can be inserted for n and p in this formula; however, the shape then is questionable. If a number of negative dots and vertices can form a geometrical shape is a complex issue, which can not be investigated at this point. Also, the shapes when p= 2 and p= 1 if these vertices lead to actual Stellar shapes is open, but it seems likely since the sequence of Tn is included in the sequence Sn.
Thus, the conclusion can be drawn that when the conditions,
n>0
p>2
therefore if p is a fixed positive integer,
(not that infinite is written out instead of ∞, this was due to technical problems)
The formula Sn=pn(n-1)+1 can be applied to derive a certain Stellar shape.
When these conditions do not apply a clear pattern of a sequence can still be observed; however, if all the values lead to definite Stellar shapes remains an open question.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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