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# Stellar numbers

Extracts from this document...

Introduction

Math Portfolio

Stellar Numbers

Triangles:

Introduction:

Throughout this task, I will first be trying to find a general statement that represents the nth triangular number in terns of n. After doing so, I will be able to complete the triangular numbers sequence by adding three more terms. I will have to go through several trial and error runs as well as numerous steps with the data that has been given in order to find the general statement, which would lead to finding the next couple of terms in the sequence.

Information Given:

Tn = number of dots

n = number of rows

 Tn n 1 1 3 2 6 3 10 4 15 5

To start off, I had to find the positive difference of Tn as well as the positive difference of n,which was simply just one as for the numbers only went up by one every time.

 Positive difference of Tn 1                        →          1 3=1+2                        →          2 6=(1+2)+3                →          3 10=(1+2+3)+4                    →          4 15=(1+2+3+4)+5        →          5

After using this information and putting it all together, it was clear to see that the polynomial had to be a quadratic equation:

Tn = an2 + bn + c

I then plugged the information given from the very start into the quadratic that I had found:

Tn = 1

n = 1

1 = a(1)2 + b(1) + c

1 = a + b + c

Tn = 3

n = 2

3 = a(2)2 + b(2) + c

3 = 4a + 2b + c

Tn = 6

n = 3

6 = a(3)2 + b(3) + c

6 = 9a + 3b + c

Middle

Sn

n

1

1

13

2

37

3

72

4

First of all, I found the number of dots (the stellar number) in each stage up to S6 by drawing each star and counting all the dots in one diagram.

 S1 = 1 S2 = 13 S4 = 72 S3 = 37 S5 = 117 S6 = 171 After counting all the dots from each diagram, I was able to put all of the information in a table.

 Stellar Shape (n) Number of Dots Increase of Dots Second Difference 1 1 - - 2 13 12 12 3 37 24 12 4 73 36 12 5 121 48 12 6 181 60 12 7 253 72 12

We can clearly observe that there is an increase of 12 dots for each consecutive stellar shape.

When I apply the pattern that was found above, I was able to find an expression for the 6-stellar number at stage S7:

S7 = 171 + (54 + (9 – 1))

S7 = 171 + 8

 S7 = 233 Following this, I was also able to find a general statement for the 6-stellar number at stage Sn in terms of n. To find it I had to go through several steps.

First, I found the positive difference of Sn:

13-1=12

37-13=24

72-37=35

After doing so, I found the positive differences of 12, 24 and 35 which were 11 and 12.Then I found the positive difference of 11 and 12 which was 1. With the information gathered, I figured out that this had to be cubic:

Tn = an3 + bn2 + cn + d

I then inserted the information given from the very start into the cubic that I had found:

Sn = 1

n = 1

1 = a(1)3 + b(1)2 + c(1) + d

Conclusion

d value would increase by 2.
 p d 6 2 7 4 8 6 9 8

5-stellar star:

p = 5

n = 2

Sn = -1/6n3 + 7n2 – 47/6n + 0

S2 = -1.33 + 28 – 15.67 + 0

S2 = -17 + 28 + 0

S2 = 11 Conclusion:

This Stellar Numbers Mathematics Portfolio required finding two general statements as well as different expressions and patterns. The first general statement that I found represented the nth triangular number in terms of n (Tn = 1/2 n2 + 1/2 n). When using this equation, the number of dots in a triangle can be calculate without having to count the dots individually.

The second general statement that was found was for the 6-stellar number at stage Sn in terms of n (Sn = -1/6n3 + 7n2 – 47/6n + 2). This equation helps calculate the number of dots in each star according to the number of rows in any stellar shape with 6 vertices. I also discovered that the d value changes depending on the number of vertices in a star.        There were also several limitations regarding the two different tasks. One was that no values could be below zero, as for that is unrealistic, and this is considered for both tasks. Another limitation regarding the first task (triangles) is that the equation only works for equilateral triangles (triangles with equal sides).  Secondly, for the Stellar Star Task, I discovered that the p value has to be at least 4 to be able to be seen in a stellar star shape. Also, the general statement for the Stellar Stars can only be used for 6-Stellar Stars.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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2. ## Maths Internal Assessment -triangular and stellar numbers

Question 8: Test the validity of the general statement Question 9 Discuss the scope and limitations of this general statement A few limitations come with the use of this general statement. There are three positive numbers that cannot be used in this general formula for the values of 'p'.

1. ## Stellar numbers

As with the 6-vertices there are two layers, the heptagon then the extensions of the heptagon. Hence, Due to their similar nature of construction in comparison to the 6-vertices stellar shape, the "n+1" shall be changed to a "n-1." Plus one is added back in, to account for the dot.

2. ## Stellar Numbers. After establishing the general formula for the triangular numbers, stellar (star) shapes ...

Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below Hence, the general formula for this sequence is Rn = 5n2 - 5n

1. ## Stellar Numbers Portfolio. The simplest example of these is square numbers, but over the ...

consistently 14, meaning that it, just like the 6-stellar and 5-stellar numbers, is based upon a quadratic general statement. I went along with the pattern and tested the validity of my first guess for the general statement of 7-stellar numbers: 7n2-7n+1.

2. ## Stellar Numbers. In this study, we analyze geometrical shapes, which lead to special numbers. ...

Using the expression one can continue the triangular number sequence indefinitely. Task 2 Find a general statement that represents the nth triangular number in terms of n. In the previous task, we defined the nth triangular number in terms of the previous triangular number In this task we can derive

1. ## Stellar numbers. This internal assessment has been written to embrace one of the ...

The 1st difference between consecutive values of yis not constant thus eliminating the possibility of the general statement being a linear one, in the form of y=mx+b. However, the 1st difference does seem to be increasing at a constant rate for each input of n.

2. ## Stellar Numbers. In this folio task, we are going to determine difference geometric shapes, ...

with dots", and it can be shown below by drawing from the computer. From the diagram, the three diagrams have black dots and white. The black dots are the original dots starts from the square number Sn=15. And the white dots are dots had added by increasing the general term number. • Over 160,000 pieces
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