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Pascal's Triangle

Dimensions

I bumped into Pascal’s Triangle whilst thinking about dimensions.

I was thinking about the minimum number of points, vertices, faces, objects that were required to construct an entity in a given number of dimensions.

For instance:
A point can exist in zero dimensions.
A line consists of two points and one vertices and exists in one dimension.
A triangle consists of three points, three vertices and one face and exists in two dimensions.
A tetrahedron consists of four points, six vertices, four faces, and has one volume.

The table below should help to clarify:

  Entity Dimensions Points Edges Faces Objects Hyper Object
Point Point 0 1 0 0 0 0
Line Line 1 2 1 0 0 0
Triangle Triangle 2 3 3 1 0 0
Tetra Tetrahedron 3 4 6 4 1 0
? Hyper Tetrahedron 4 5 10 10 5 1

Notice that with the exclusion of the ‘Dimensions’ column, that the value of any number square that you pick is the sum of the number in the square above and the square above and to the left.

The numbers in the table above which are bold, form half of Pascal’s triangle. I was amazed when I realised this!

Hyper Tetrahedron

Using the table above, reading the values from the ‘Hyper Tetrahedron’ row, we can predict that a four dimensional 'Hyper Tetrahedron' would have 5 points, have 10 edges, 10 faces and have 5 objects!

It is rather difficult to visualise a 'Hyper Tetrahedron' because in our world we seem only able to perceive entities only with a maximum of three dimensions (excluding time).

The Hyper Tetrahedron is also known as a ‘Pentatope’ and as a ‘Simplex’.

Here is another image I rendered to help in the visualisation.

InfiniteDimensions3

Pascal's Triangle

Pascal’s Triangle is pictured below:
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
1
5
10
10
5
1
1
6
15
20
15
6
1
1
7
21
35
35
21
7
1
1
8
28
56
70
56
28
8
1


Number Patterns

Pascal’s Triangle contains many amazing patterns:

Powers of Two

1) The sum on the numbers in each row is equal to a power of 2.
1 =1 =20
1+1 =2 =21
1+2+1 =4 =22
1+3+3+1 =8 =23
1+4+6+4+1 =16 =24
1+5+10+10+5+1 =32 =25
1+6+15+20+15+6+1 =64 =26
1+7+21+35+35+21+7+1 =128 =27
1+8+28+56+70+56+28+8+1 =256 =28

Powers of Eleven

2) The sequence of numbers in each or the first 5 rows is equal 11 raised to a power.
1 110 1
11 111 11
121 112 121
1331 113 1331
14641 114 14641

Fibonacci

3) Adding the numbers in Pascal’s Triangle reveals the Fibonacci sequence.Pascal-Fibonacci


Triangular Numbers

4) The ‘Triangular’ numbers are here. Pascals -Triangular

The ‘Triangular’ numbers may be visualised by examining the table below. As can be seen they are effectively (1), (1+2), (1+2+3), (1+2+3+4) etc. To calculate the nth ‘Triangular’ number, see Sum of Series.

1 3 6 10 15 21 28
Triangular-Number-1 Triangular-Number-3 Triangular-Number-6 Triangular-Number-10 Triangular-Number-15 Triangular-Number-21 Triangular-Number-28

Square Numbers

5) Summing pairs of numbers reveals the ‘Square’ numbers. Pascal-Square-Numbers The ‘Square’ numbers may be visualised by examining the table below.

Notice that the ‘Square’ number n is the products of the sum of the ‘Triangular’ numbers n + n-1  i.e. (0+1), (1+3), (3+6), (6+10), (10+15) etc.

It is interesting to notice that nth ‘Square’ number is also the sum of the series of n odd numbers. i.e. (1), (1+3), (1+3+5), (1+3+5+7) etc.

1 4 9 16 25 36 49
>Square-Number-1 Square-Number-4 Square-Number-9 Square-Number-16 Square-Number-25 Square-Number-36 Square-Number-49
 

Sierpinski Triangle

6) If you colour the odd and even numbers in Pascal’s triangle you get this. Pascal-Sierpinski Here is a Pascal’s triangle with a few more rows, so you can see the pattern emerging. Pascal-Modulus2 The pattern produced is very similar to the Sierpinski triangle.

Binomial Theorem

The number sequences in the rows of Pascal’s Triangle are used as the coefficients in Binomial Theorem.
Click on an Image

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