MATH IB SL INT ASS1 - Pascal's Triangle

a=0.5

b=0.5

c=0

Therefore:

f(x) = 0.5x² + 0.5x + 0

As one can see, this is the same formula as we calculated before.

Afterwards I tested the validity of the general formula by calculating some y-values which represent Xn. The x-values which I inserted in the general formula represent n. Furthermore I created a table with the calculated y-values to compare it with the table above (Fig. 2).

Figure 5: Xn calculated with the general formula

When we compare the two tables (Fig. 2 and Fig. 5) we can state that the values for Xn are the same. Therefore we can consider the general explicit formula for Xn as valid.

The advantage of this formula is obvious, because we are able to calculate any given numerator of any row, without knowing the previous one.

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Find the sixth and the seventh rows. Describe any pattern you used.

6th row:        

7th row:        

For the calculation of the numerator I used the pattern which I already described above (see page 1):

Xn = n + Xn-1

For the 7th row:

Xn = 7 + 21 = 28

For the denominator I used a different pattern. The general approach is as follows:

Let the denominator of each fraction within a row be Yn(r), whereby n is the number of the row and r is the position of the fraction within the row, starting with r = 0.

Moreover, I defined some diagonals within the triangle for orientation. Each diagonal has the name dr, because every fraction which is on this diagonal has the position r:

d4

d3

d2

d1

d0

There is a clear pattern throughout each diagonal how the number of the denominator increases.

If we have a closer look at d1, for example, we can easily determine this pattern:

The first number is 1, also can be written as

 and the denominator is therefore 1.

The second fraction is

 and the denominator is therefore 2.

The third fraction is

 and the denominator is therefore 4.

If we continue this procedure and put then the denominator of each fraction in a series it would look like 1; 2; 4; 7; 11; 16; … .

Apparently, if we want to calculate a denominator of a fraction in d1  in the nth row, we always have to add (n-1) to Yn-1(1), or, to apply in a formula:

Yn(1) = Yn-1(1) + (n-1)

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If we do the same with d2 we will get a series of 3; 4; 6; 9; … .

The formula would be:

Yn(2) = Yn-1(2) + (n-2)

If we do the same with d3 we will get a series of 6; 7; 9; 12; 15; … .

The formula would be:

Yn(3) = Yn-1(3) + (n-3)

The question why the last two series start with 3(6) will be discussed later.

The formulas mentioned above make another pattern visible:

We always have to subtract r from n, which leads us to the general recursive formula for the denominator:

Yn(r) = Yn-1(r) + (n-r)

With this formula we can calculate any denominator of the triangle provided if we know Yn-1.

However, there is one diagonal which does not seem to fit into this pattern. It is the diagonal d0, because in this diagonal occurs only number 1. Nevertheless we can transfer our pattern even on this diagonal:

Let a fraction in the nth row with the position r within the row be En(r), starting with r=0.

Let us assume that Y1(0) = 1 = X1 =

 = E1(0) =

E2(0) can therefore also be written as a fraction, which is 1 =

, because

Yn(r) = Yn-1(r) + (n-r)

Y2(0) = Y2-1(0) + (2-0)

Y2(0) = 1 + (2-0) = 1+2 = 3

therefore Y2(0) = 3

and

Xn = n + Xn-1

X2 = 2 + 1 = 3

therefore X2 = 3.

E2(0) =

 =

 = 1

Apparently, we are able to determine the denominator even in d0.

This why the series of d2 and d3 starts with 3(6), because Yn(0) = Yn(n).

Moreover we see that the denominators in d0 are equal to the numerators, so Xn = Yn(0).

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However, we will not get a solution if we try to calculate Yn(n) because there is no

Yn-1(n). So we have to assume that Yn(n) = Yn(0).

If we, for example, want to calculate the last denominator of the 4th row, Y4(4), where n=4, we have to insert in our formula:

Y4(4) = Y3(4) + (4-4)

However, we cannot calculate Y4(4) because Y3(4) is not defined in the triangle.

Therefore we assume Xn = Yn(0) = Yn(n)

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Find the general statement for En(r):

We already have determined the general recursive formula for the numerator and the denominator.

Xn = n + Xn-1

Yn(r) = Yn-1(r) + (n-r)

If so

En(r) =

 =

 =

However, this formula has some disadvantages:

First, we are not able to determine Yn(n) because there is no Yn-1(n), although we know that

Yn(n) = Yn(0) = Xn (see also page 7).

Second, it is a recursive formula, thus we always have to know Xn-1 and Yn-1(r). This would not work properly if n and r are large numbers. Moreover, calculation would require a long period of time.

Therefore we need an explicit formula.

We actually know the explicit formula for the numerator which is:

Xn = 0.5n² + 0.5n + 0.

In order to find the explicit formula of the denominator I used the same strategy as for the numerator.

First of all, I plotted the denominators of the 5th row up to the 9th in a table. These numbers were calculated by using the recursive formula for Yn(r).

Figure 6: The denominators of some rows plotted in a table.

We can conclude from the table, that the values of the denominators must be arranged in a parabolic form, because there is a clear minimum in each row and each number occurs two times in the same pattern: the first number equals the last number, the second number equals the second last number and so on.

To check if this is true, I plotted the data in a diagram.

Figure 7: Some values for Yn(r) shown in a diagram.

Obviously, the denominators are ordered in a parabolic form. Therefore it should be possible to find a line of best fit for each row. Furthermore, it should be possible to find a general pattern within the formulas for these lines. This should lead us to a general formula, which describes the data shown in Figure 7.

In order to do so, I calculated a line of best fit for each row with a GTR (Fig. 8).

Figure 8: Several functions which describes the denominators of the different rows within the triangle

I got the following formulas for the different functions:

For Y5(r): f(x) = x² - 5x + 15

For Y6(r): f(x) = x² - 6x + 21

For Y7(r): f(x) = x² - 7x + 28

For Y8(r): f(x) = x² - 8x + 36

For Y6(r): f(x) = x² - 9x + 45

Within these functions there is a clear pattern:

If we take the general formula of a parable f(x) = ax² + bx + c we can easily compare our functions with the general one.

The parameter a in our functions is always 1.

The parameter b in our functions is always -n, thus we can write b=-n.

The parameter c in our functions is always the number of the numerator, thus we can say c=Xn or c = 0.5n² + 0.5 (which is the general formula for Xn) .

The parameter x represents the position of the fraction within the row, thus we can say x=r.

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Therefore, we can write a general formula for Yn(r):

Yn(r) = r² - nr + Xn                                 or

Yn(r) = r² - nr + (0.5n² + 0.5n)

This is the explicit formula for the denominator.

To test the validity of the statement, I calculated the denominators of the 5th up to the 9th  row with the general formula, to compare the data with the table above (Fig. 6).

Figure 9: Showing the denominators, calculated with the general formula.

As we can see, the calculated numbers for the denominator in the table above are the same as in Figure 6.

Therefore we can consider the general formula for the denominator as valid.

To get the explicit formula of En(r) we only have to put the explicit formulas for the numerator and the denominator together:

En(r) =

 =

Xn = 0.5n² + 0.5n

Yn(r) = r² - nr + (0.5n² + 0.5)

En(r) =

This is the general formula for En(r).

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We have already tested the validity of the general explicit formulas for the numerator and the denominator and we regarded them as valid. Therefore the general formula for En(r) has to be valid.

Nevertheless I tested the validity of the general formula for En(r) one more time.

We know that the fractions of the 8th row are

 and the

fractions of the 9th row are:

.

I have calculated the fractions of these rows once more, by using the general explicit formula for En(r).

The result for the 8th row was:

The result for the 9th row was:

.

Obviously, the general formula for En(r) leads us to the correct results, therefore we can regard the formula as valid.

Discuss the scope and/or limitations of your general statement:

Compared to the recursive formula, the explicit one has several advantages:

The major one is, that we are able to calculate the fractions of any given row, without knowing the previous one.

However, we have never proven that the general statement is true, we just checked it for a few examples and not for all rows (which would be for n=∞). Theoretically it could be coincidence that we got right result with this general formula.

Moreover, the formula does not only describe the fractions of a existing row, but also the fractions if n would be a decimal number (for example n=1,34), although this row does not exist in the triangle.

Beyond this, if we calculate the fraction with a calculator, we will not get the fractions as they are in the triangle (which are not cancelled). We will either get a decimal number or a cancelled fraction. Hence, if we want to calculate the fractions as they are in the triangle, we have to calculate numerator and denominator separately.

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