Introduction to Pascal's Triangle

In this essay I'd like to introduce some clever ways of doing algebra and of calculations using a simple calculator. By simple calculator I mean one that does only the basics (plus, minus, multiply and divide). When you have finished this essay you should be able to calculate quite difficult roots with this calculator. But before picking up a calculator, let us look at Mr Pascal.

Blaise Pascal (1623 - 1662) was a French mathematician. His surname is used as the unit of pressure. One of his sayings was to note that 'had Cleopatra's nose been differently shaped, the history of the world would have been different'. He is most famous for the triangle named after him, Pascal's Triangle. It's not a geometrical triangle but a triangle of numbers. Here it is below:

Study these numbers and see if you can figure what the next line should be before reading on . . .

Each number in the triangle is the sum of two above. For example, the 6 on line 5 is the sum of the pair of 3's above. So the next line is

1, 10 (1 + 9), 45 (9 + 36), 120 (36 + 84), etc.

I am now apparently changing the subject and turning to a bit of algebra.

Algebraic Expansions

Just to remind you:

We can do the same with a cube, i.e (1 + x)³.

If you look at the coefficients (the numbers on their own and in front of the x's) of the results you will see that for the first one they are 1, 2, 1 and for the second one they are 1, 3, 3, 1. These, of course, are the lines from Pascal's Triangle. And yes, it does work for all positive whole number values of the index. Prove to yourself by algebra that,

In fact there is a general rule that

As you see, the indexes begin at 4 and descend for the a's while they ascend to 4 for the b's. The coefficients are the 1, 4, 6, 4, 1 from Pascal's triangle. So let's try an example. Expand (2 + 3x)⁵.

Applying the general rule of ascending and descending indexes and the coefficients from Pascal we can immediately expand the above equation (if we set a to 2 and b to 3):

This simplifies to

It is now time to apparently change the subject again...

Selections and Combinations

How many selections are there if we wanted to select two books? Well, let's list all the combinations:

That makes 10 selections.

What about if we wanted three from five? Well, that's easy. Picking three from five is the same as discarding two from five so there are ten ways of doing this. Picking four from five is the same as discarding one from five, five ways. Of course, if you wanted to select all five books there is only one way of doing that. There is also just one way of selecting no books! So tabulating we have:

Again, these numbers are a line from Pascal's Triangle.

This is a fascinating aspect of mathematics that two seemingly unconnected bits are in fact related. Algebraic expansions and selections.

The number of ways of selecting r objects from a total of n is written as

So the above selections can be written mathematically as follows:

So, if somebody asks you 'how many ways are there of selecting six objects from eight?'. You can say, ⁸C₆. When they look at you blankly you can pull out your copy of Pascal's Triangle, that you carry everywhere, and look it up. There are total of eight objects so you look at the line beginning with 1, 8, etc. You want to select six so you count along from zero, until you count six. The number there is 28 so there are 28 selections.

Mathematically we can write ⁸C₆ = 28.

Rather than having to carry a copy of Pascal's Triangle, it would be nice if there was a formula for calculating ⁿC_r. It could be used both for selections and for expanding algebraic expressions. Remember that

There is a formula and it is very easy to use but looks difficult. So before I tell it to you, relax and re-read what we have covered up to now.

Formula for ⁿC_r

Now, before you panic, n! (pronounced n factorial) simply means multiply together all the whole numbers from 1 to n. So,

Also, and don't ask me why, 0! = 1.

Yes, it looks silly but accept it for now. So let's go back to our selections. How many ways can you select 8 books from 11? The answer is, of course ¹¹C₈ which is given by:

This is now a case of cancelling down to 165. So there are 165 ways of selecting 8 books from 11.

Using the above formula instead of Pascal's Triangle we can look at one of our expansions and say that

and show that

³C₀ = 3! / 3! 0! = 1 x 2 x 3 / 1 x 2 x 3 x 1 = 1 (remember 0! = 1)

³C₁ = 3! / 2! 1! = 1 x 2 x 3 / 1 x 2 x 1 = 3

³C₂ = 3! / 1! 2! = 1 x 2 x 3 / 1 x 1 x 2 = 3

³C₃ = 3! / 0! 3! = 1 x 2 x 3 / 1 x 1 x 2 x 3 = 1.

Thus the numbers of Pascal's Triangle can be calculated from the formula above for ⁿC_r. We can do this for any value of n that is a positive whole number.

In summary, Pascal's Triangle can be used:

• to form the basis of selection probabilities (these are called Combinations, hence the C in the ⁿC_r above).
• to determine the coefficients of the expansions of (1 + x)ⁿ, where n is a positive whole number.

Just to remind you, when n is a positive whole number the expansion contains x + 1 terms.

The expansion for (1 + x)³ contains 4 terms.

The Binomial Theorem

Isaac Newton, as we all know, did a lot of good work with apples. He was also a brilliant mathematician, having to invent things like calculus for himself so that he could work out his equations on gravity.

He looked at the formula for the expansion of (1 + x)ⁿ and thought 'Why does this only work for positive whole values of n?' Since there was nobody around to answer that question he did it himself.

He decided that there was an expansion for (1 + x)ⁿ that could be derived from the formula for ⁿC_r that worked for all values of n (fractions, negatives, etc). Before we look at what fractional and negative indexes actually mean I'll write down Newton's formula.

This series is called the Binomial Series.

When n is a positive whole number the series has n + 1 terms and produces the same results as before. However when n is not a positive whole number then the series goes on for ever. This is called an infinite series.

There are two types of infinite series. Imagine an infinite series that goes like this:

A series like this gets bigger the more terms you add. If you took it all the way to infinity, then the sum of the series would be infinite. This is called a diverging series. This kind of series is not of much use for anything.

Now look at this infinite series:

This series also is infinite, it too goes on forever. However, the more terms you take the smaller each one gets. This series never goes above 2. As you take more and more terms the sum gets closer to 2. This kind of series is said to be converging.

The nice thing about converging series is that they can be used to do calculations. You take as many terms as are needed to make the calculation as accurate as you require. In the above example, taking the first five terms gives you an accuracy of 2 decimal places.

In the Binomial Series, Newton discovered that for values of n that were not positive whole numbers (i.e. for fractional and negative indexes), the series converges only if the value of x is 1 or less and more that -1. In symbols, the series converges for

For all other values of x the series diverges. In other words there are specific situations when the Binomial Series can be used for calculations of approximations.

Let's do a calculation by first using Newton's Binomial formula to expand
(1 + x)^1/2. If you're patient I'll soon explain what an index of 1/2 means.

Doing a bit of simplifying algebra gives us:

So what does an index of 1/2 mean? Well, without going into details, x^1/2 is the square root of x (Öx).

So, the formula above, the expansion of (1 + x)^1/2, can be used to calculate approximate values for square roots as long as x is less that or equal to 1.

If we let x = 1 then this formula will give us a value for the square root of 2, since
2^1/2 = (1 + 1)^1/2 and this can be expanded into the series above. If we do the calculation up to and including terms with x³ then we get

This equals 1.4375. The calculator answer is 1.4142. The more terms you use the closer the approximation gets to the real answer. Notice also how each term is smaller then the previous one.

Let's do another.

Find an approximation for the square root of 1.77.

Start by saying 1.77^1/2 = (1 + 0.77)^1/2 which can be expanded by using the Binomial Theorem to give:

which gives 1.3394 (calculator 1.3304). You see this is accurate to two decimal places.

Wonderful, you may say, but what if you want the square root of, say, 30?
You can't write 30^1/2 = (1 + 29)^1/2 because the Binomial Series does not converge if x > 1.

But I'm going to show you how.

First you express 30 in the form that includes the largest perfect square.

Instead of saying 30^1/2 = (1 + 29)^1/2 which doesn't work, we write:

25 is the largest perfect square below 30. We can take out the 25 (remembering the index) and divide everything inside the bracket by 25.

This gives us the following:

Since 25^1/2 is 5 (the square root of 25), we can rewrite this expression as:

The bit inside the bracket is now in the form (1 + x) with x < 1 so we can use Newton's Binomial expansion to get a value for the square root of 1.2. We then multiply this value by 5 (the number outside the bracket). This will give us the square root of 30. Doing the calculation we get:

The calculator says 5.4772 (so we are accurate to three places!). Try it yourself. Remember the trick is to write the number as a perfect square plus or minus another number.

To find the square root of 45 we would write it as

This gives

Cube roots can also be done this way! After all, if an index of 1/2 is a square root, it follows that an index of 1/3 is a cube root. In fact an index of 1/n is the nth root. Expanding Newton's formula for (1 + x)^1/3 gives us the formula for cube roots.

which simplifies to:

This can be used to find cube roots. Let's find the cube root of 30. Like last time we need to write it as the sum of two numbers. This time one of the numbers must be a perfect cube.

We can say 30^1/3 = (27 + 3)^1/3 because 27 is a perfect cube (3 x 3 x 3 = 27). By taking out the 27^1/3 and using Newton's formula we can write:

This now expands (using the formula above) to:

The calculator says 3.10723, so again we are correct to two decimals.

Logarithms can also be used to calculate roots.

I'll end here with a little something for you to find out.

If 2² = 2 x 2, 2³ = 2 x 2 x 2 and 2^1/2 is the square root of 2, what do 2^1.5, 2⁰ (not the obvious answer), and 2^-1 mean?

Here are the answers and there is more on the properties of indices (plural of index) in the essay about Logarithms.

© 1998 Kryss Katsiavriades

Related Pages

Introduction To Algebra
An introduction to algebra and how to solve simple, simultaneous and Quadratic equations.

Trigonometry
Right-angled triangles, Sines, Cosines, Tangents. Using Trigonometric Functions, series and formulas.

Look At Logarithms
More on indices and series. Calculations of multiplication, division, roots and powers using logarithms.

 

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Answers

2^1.5 = 2^3/2 = Ö2³ = Ö8

2⁰ can be obtained from 2ⁿ / 2ⁿ = 2^{n - n} = 2⁰.

but 2ⁿ / 2ⁿ = 1

therefore 2⁰ = 1.

In fact, any number raised to the power of 0 is 1.

2^-1 can be obtained from (say) 2² / 2³ = 2^{2 - 3} = 2^-1.

but 2² / 2³ = 1 / 2

therefore 2^-1 = 1 / 2.

In fact, any number raised to the power of -1 is the reciprocal of the number.