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Pascal's Trianglecombinations : algebra : probability : binomial theorem : roots 

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). At the end, you should be able to calculate quite difficult roots with this calculator.
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. In fact, the triangle was known to both the Chinese and the Arabs for several hundred years previously.
It is 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.
By "expanding" I mean remove the brackets. As a reminder, the expression is expanded like this:
In the middle pair of brackets, each term in the left bracket is multiplied by each term in the right bracket. We can do the same with a cube, i.e (1 + x)^{3}.
= 1 + 3x + 3x^{2} + 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...
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:
Number of selections from 5  0  1  2  3  4  5 
Number of ways of making selections  1  5  10  10  5  1 
Again, these numbers are a line from Pascal's Triangle.
This is a fascinating aspect of mathematics that two seemingly unconnected topics are in fact connected. In this case we have a connection between 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:
There is one way to select no books from 5  ^{5}C_{0} = 1 
There are 5 ways of selecting 1 book from 5  ^{5}C_{1} = 5 
There are 10 ways of selecting 2 books from 5  ^{5}C_{2} = 10 
There are 10 ways of selecting 3 books from 5  ^{5}C_{3} = 10 
There are 5 ways of selecting 4 books from 5  ^{5}C_{4} = 5 
There is one way to select 5 books from 5  ^{5}C_{5} = 1 
Question: "How many ways are there of selecting six objects from eight?". Mathematically, the answer is^{8}C_{6}. From Pascal's Triangle, 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 ^{8}C_{6} = 28.
Rather than having to memorise Pascal's Triangle, it would be useful if there was a formula for calculating ^{n}C_{r}. It could be used both for selections and for expanding algebraic expressions. Remember that
= 1 + 2x + x^{2}.
There is a formula and it is very easy to use but looks difficult. It may be useful to reread the first part of this essay again.
The term n! (pronounced n factorial) means multiply together all the whole numbers from 1 to n. So,
1! = 1 
2! = 1 x 2 = 2 
3! = 1 x 2 x 3 = 6 
4! = 1 x 2 x 3 x 4 = 24 
5! = 1 x 2 x 3 x 4 x 5 = 120 
Also, 0! is defined as 1.
It is now time to return to selections. How many ways can 8 books be selected from 11? The answer is, of course ^{11}C_{8} which is given by:
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 / 1 x 2 x 3 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8
= 9 x 10 x 11 / 1 x 2 x 3
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
^{3}C_{1} = 3! / 2! 1! = 1 x 2 x 3 / 1 x 2 x 1 = 3
^{3}C_{2} = 3! / 1! 2! = 1 x 2 x 3 / 1 x 1 x 2 = 3
^{3}C_{3} = 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 ^{n}C_{r}. We can do this for any value of n that is a positive whole number.
In summary, Pascal's Triangle can be used:
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.
He decided that there was an expansion for (1 + x)^{n} that could be derived from the formula for ^{n}C_{r} that worked for all values of n (fractions, negatives, etc). Before we look at what fractional and negative indexes actually mean I will 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 useful 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 than 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
I will 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 (√x).
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 than 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 x^{3} 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 us 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:
= 1 + 0.385  0.0741 + 0.0285  ...
which gives 1.3394 (calculator 1.3304). You see this is accurate to two decimal places.
This is all well and good, but what if the square root of a larger number (like 30) is wanted?
You cannot write 30^{1/2} = (1 + 29)^{1/2} because the
Binomial Series does not converge if x > 1.
There is a way around that.
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 term 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 us find the cube root of 30. Like previously, 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.
As a matter of interest, Logarithms can also be used to calculate roots.
I will end here with a little something for the reader to find out.
Here are the answers and there is more on the properties of indices (plural of index) in the essay about Logarithms.
© 1998, 2009 KryssTal
2^{0} can be obtained from 2^{n} / 2^{n} = 2^{n  n} = 2^{0}.
but 2^{n} / 2^{n} = 1
therefore 2^{0} = 1.
In fact, any number raised to the power of 0 is 1.
2^{1} can be obtained from (say) 2^{2} / 2^{3} = 2^{2  3} = 2^{1}.
but 2^{2} / 2^{3} = 1 / 2
therefore 2^{1} = 1 / 2.
In fact, any number raised to the power of 1 is the reciprocal of the number.