## Pascal Triangle Essay

## Introduction to Pascal's Triangle

Mathematics is the language of science. Something that may be difficult to picture may be easy to understand mathematically. A mathematical equation may take up a single line whereas the same thing written in words may take up a large paragraph.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.

## Algebraic Expansions

Imagine having to expand an expression likeBy "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}.

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

Imagine you have five books. Let us imagine that you wanted to read one. How many ways are there of selecting a single book? Well, that is easy, there are 5 ways. If we label the books A, B, C, D, E, we can chose any one of five; five different selections. How many selections are there if we wanted to select **two** books? Well, let's list all the combinations:

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:

**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:

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

## Formula for ^{n}C_{r}

The formula for ^{n}C

_{r}is:

The term **n!** (pronounced **n factorial**) means multiply together all the whole numbers from 1 to n. So,

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:

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

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:

**to form the basis of selection probabilities (these are called****Combinations**, hence the C in the^{n}C_{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

During the 10th century, various Arab mathematicians developed a mathematical series for calculating the coeficients for (1 + x)^{n}when n was a positive whole number. The English mathematician,

**Isaac Newton**extended it to non-integer indices in the 17th century.

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.

**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**.

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.

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:

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**

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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.## Answers

**2**

^{1.5}= 2^{3/2}= √2^{3}= √8**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.

## Essay on Pascal’s Wager

984 Words4 Pages

In this paper I will be discussing Pascal’s Wager. What I first plan to do in this paper is explain the argument of Pascal’s Wager. Next I will explain how Pascal tries to convince non-theists why they should believe in God. I will then explain two criticisms in response to Pascal’s argument. Finally, I will discuss whether or not these criticisms show Pascal’s reasoning to be untenable.

Pascal’s Wager is an argument that tries to convince non-theists why they should believe in the existence of the Christian god. Pascal thinks non-theists should believe in God’s existence because if a non-theist is wrong about the existence of God they have much more to lose than if a theist is wrong about the existence of God.

Pascal begins his*…show more content…*

If this is true then the non-theist will be rewarded in this life and will not be rewarded or punished when they die. Since there are only two choices a person can make Pascal believes a person should choose to believe in God. Pascal comes to this conclusion based on what he believes are the four possible outcomes of a person’s choice. Pascal believes believing in God is the best possible choice because between placing a wager on God’s existence and placing one on God’s non-existence, placing a wager on his existence offers the best rewards. Pascal explains that if a believer is wrong about God’s existence then they only suffered while they are alive, however, if a nonbeliever is wrong about God’s existence then while they did enjoy their life on Earth they will suffer forever in the afterlife. Since the possible rewards are greater than the possible suffering that a theist can experience it is in everyone’s best interest to believe in God’s existence.

While Pascal’s Wager is a somewhat convincing argument to become a Christian theist this argument has two important criticisms. The first criticism I would like to bring up is one Pascal refutes himself in the Pensees. The second criticism I would like to bring up is known as the “many-gods objection” this argument Pascal does not even acknowledge.

Pascal’s Wager has been argued to be impractical because our beliefs are often not in our control. This argument is

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