Patterns are a series of discrete numbers. This means that even though most patterns can be represented by a function, the pattern would be represented on a graph by individual points and not a continuous line.
Most patterns can be represented by a function. Some are harder to find that others, but you might want to look for regular changes from one element of the list to the next.
An example of such lists and how to find the pattern include:
List:
1 3 5 7 9
11 …
\ /\ /\ /\ /\ /
Differences: 2 2
2 2 2
Function: There is a first level change of two per element, so we know that we add 2 times some value based on the element number. Let us initially try y = 2x. If x = 1, then this would imply that the first element is 2. This is one too high, so let us try y = 2x –1. Now try the second element, does 3 = 2*2 – 1. Yes. And 2*4 –1 = 7, so this appears to fit for all elements we have been given.
Not all functions are linear. For example, the initial patterns in chapter 4-1 represent starting with an initial set of five squares and then adding one set of squares around this set and continuing on as long as is desired. If we take the total number of squares included in the entire figure as our pattern, then the total number of squares would be:
List: 5 13
25 41 61 75 …
\ /\ /\ /\ /\
/
Differences: 8 12
16 20 24
\ / \ / \ / \
/
Second Level: 4 4
4 4
Function: This one is a bit tougher to calculate. There is a second level change of 4 per element, so it is not as easy as the first example. If you were in Pre-Calculus, then you might write a function like:
For x equals 1 to n, sum 4 * x and then add 1.
For example: If n = 5, the sum would be: 5*4 + 4*4 + 3*4 + 2*4 + 1*4 + 1 = 20+16+12+8+4+1=61
This may be the correct way, but it might also not be the fastest way. You might want to look for patterns that you might not expect. For example, I just tried squaring the pattern index and compared it to the pattern number. I got:
|
Index |
Pattern |
Index2 |
Difference |
|
1 |
5 |
1 |
4 |
|
2 |
13 |
4 |
9 |
|
3 |
25 |
9 |
16 |
|
4 |
41 |
16 |
25 |
|
5 |
61 |
25 |
36 |
|
6 |
75 |
36 |
49 |
The interesting observation is that the difference is (x+1) 2. So, I could now write the formula as:
f(x) = x2 + (x+1) 2 or x2 + x2 + 2x + 1 or 2x2 + 2x + 1
If you want to go back to your diagram and see why this is the case, it might be a fun exercise..
1 2 3 4 5 …
2, 3, 4, 5, 6
or
1 1
1 1 1
…
2, 3, 4, 5, 6
It is nice to know. See
the linked web-page.