When multiply out this bracket we get all the combinations of powers for a and b such that the sum of their powers is n. So there is a term of aⁿwhere the power of b is 0 and one of bⁿ and then there are terms in between with both a and b raised to powers. Then all that is left is to find the co-efficents of each of these terms. For the case that n is a positive integer there are two ways of doing this, one is to use the Binomial Theorem or the other is to use Pascal’s triangle.

Binomial Theorem

The binomial Theorem is:

Example

If we know calculate the factorials this becomes

Which you can check since we know 3+1 = 4 and 4⁴ = 256.

However there is another simpler way of finding the coefficients for the power which is to use a Pascal’s triangle. This is an quick to calculate way for moderately large powers (the triangle can get quite large for big powers above say 10) by hand. A Pascal’s triangle with 5 rows is below.

Each item in the triangle is the sum of the two numbers above it and there is a 1 at the end of each row. When the binomial terms are written in order of powers of x (either way as it is symmetric) each number in the nth row is the coefficient of x in corresponding term (check it with the example above)

More complex sums

We can also use the binomial theorem on sums where the first term is something other than 1. In general when we want to find (a+b)ⁿ.
To do this we first factor out a within the sum to get

We then treat as x and use the binomial theorem to expand to get

we can then multiply through the sum by aⁿ to get

Example

Expand (3 + 2x)⁴.

Firstly factorise the 3 to get

Using Pascal’s triangle or our previous calculations we can find the coefficients and write out the sum

and now working out the arithmetic of each term we get our final answer

The binomial theorem can also be used to calculated sums raised to non integer or negative powers but that is for another post.

Also see

• Log Laws: Taking logs of powers (0)
• Why anything to the power 0 is eqaul to 1 (2)

1 + 2 + 4 + 8 + 16 + …       each term is twice the previous one
4 + 12 + 36 + 108 + …       each term is three times the previous one

In general the nth term of a geometric series is written as where a is the first term and r is the factor we multiply each term by. If a is 1 then the terms are just the powers of r.

We are often interested at find the value of a geometric sum, that is the sum of all the values of a geometric series to a certain point. We can also find the value of the sum of infinitely many terms if . To find both of these first let the sum of the first n terms be denoted S_n.

So

The formula for a finite sum that we will try to prove is

Proof of finite sum

Consider
from the definition of S_n


We can now factorise the left-hand side and tidy up the right multiply the r through the first sum and change the index down on the second,





We can now break up the sums into parts in order that we cancel bits of them in the next step





So cancelling out the two sums and factorising the a noting that anything the the power of 0 is 1 gives





and then dividing through by (r-1) gives us our required result


.

Infinite Sums

As noted earlier is we can also find the value of an infinite sum. It may seem like a strange idea to be able to find the value of the sum of an infinite number of things but if the size of each term keeps getting smaller so the amount the sum increases by each time gets smaller and smaller.

A simple way of visualising how this can work is to imagine a piece of rope two meters long. We are going to cut this rope up into an infinite number of pieces with the length of each piece representing a term in the sum. Firstly cut the rope in two. Put one piece to the side as this is our first term so a=1 as this piece is 1 meter long. r=1/2 because each piece of rope will be half the length of the previous one. Now cut the remaining piece of the rope in half and again put one piece to the side as the second term and cut the remainder in half. Theoretically we will always be able to repeat this process because you can always halve something. This will mean that we will now have an infinite number of pieces of rope with lengths 1m, 1/2m, 1/4m, 1/8m, … so the lengths of the pieces of rope form a geometric progression and we know that the sum of their lengths must be less than two as we only started out with two meters of rope. In fact it is exactly two which we can prove.

In general,


if


To show this simply note that rⁿ tends to 0 as n tends to infinity. Then using the algebra of limits and multiplying the top and bottom y -1 we find the above expression to be true.

Also see

• Arithmetic Series and Progression (2)
• Proof there are infinitley many prime numbers (0)

Examples

Some simple examples which contain only one variable are:

Solving and Manipulating Inequalities

Inequalities can be solved by rearranging them and isolating the variable you want to find in a similar way to normal equation (see the post quadratic inequalities to see how to solve quadratics). However, rather than getting an exact value such as x=3 we get a range (open or closed) of values such as x<2 or -3<-1.

Much of the manipulation is the same though there are slight variations when dividing or multiplying by negative numbers or taking the reciprocal. The important thing to remember is that like normal equations we must do the same to both sides.

Addition and Subtraction

Addition and subtraction are exactly the same to equalities. We can add or subtract whatever we like as long as we do the same to both the sides. This enables us to take expressions “to the other side” by reversing their sign. For example all the following manipulations are valid.

Multiplication and Division

Again we can perform multiplication and division in a similar way to the way we perform it with equalities by doing the same to both sides. However, if we are multiplying or dividing by a negative number we must reverse the direction of the inequality since

This means we must be careful when diving by an unknown since by definition we don’t know whether or not it is positive or negative. If this has to be done you should consider both the cases it is positive and negative separately and if it is only positive or negative then the other inequality should lead to a contradiction which can easily be spotted such as x<0 and x>3.

Examples of valid manipulation are below:


for and/or for

Reciprocals

When taking the reciprocal or “one over” of an expression you must reverse the inequality so

Also see

• Simplifying and Solving equations with Algebraic Fractions (0)
• Simultaneous Equations (0)
• First Order Differential Equations (6)

the equation

has no integer (whole number) solutions for n>2

For example a solution for n=2 is x=3, y=4, z = 5 since

however the theorem states that for any n larger than 2 a set of integer solutions such as these cannot be found.

Despite the simplicity of this theorem it took 300 years until 1994 for it to be solved by Andrew Wiles using advanced maths.

A fascinating book on the problem is called Fermat’s Last Theorem (by Simon Singh) which goes through the history of the problem and many of the people who have attempted to solve it.

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This can be shown using De Moivres theorem. Firstly consider the complex number z = cosΘ + isinΘ and let

zⁿ = 1
(cosΘ+isinΘ)ⁿ = 1

which using De Moivres theorem gives

cos nΘ + isin nΘ = 1

We can now compare the real and imaginary parts to find the values of n

cos nΘ = 1 and sin nΘ = 0

Therefore nΘ = 2pπ for integral values of p
So

which gives

so the roots of unity are



etc

However what is interesting about these is when you plot them on an argand diagram they are evenly spaced around a circle of radius 1 and centred at the origin. An example for n=8 is shown below

The roots of unity are space evenly around a circle of radius 1 and centre at the origin

This happens because the increase in the angle for each successive root is equal since we divided 2pi by n.

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1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,
46368,75025,121393,196418,317811,514229,832040,1346269….

which quickly become very large.

The Fibonacci numbers, f_r can be defined by
f₁ = f₂ = 1
and
f_k = f_k-1 + f_k-2 for k≥3

PHI the golden ratio

PHI or Φ is said to be the golden ratio since so many things in nature seem to naturally arrange themselves in this ratio. It is approximately equal to 1.618033988749895.

Phi is also the positive solution to the equation

x² = 1 + x

which has the irrational solution

phi = (1+√5)/2

Interestingly an irrational number is one which cant be written as the ratio of two integers so the golden ratio is not in fact a ratio meaning ratios in nature can only become very close to it but cant actually equal it.

PHI and Division of Fibonacci numbers

It is interesting to calculate the ratio of consecutive numbers in the Fibonacci sequence. These ratios quickly approach the golden ratio know as PHI or Φ.

The start of list of value obtained from these divisions are:
1
2
1.5
1.6666666666666667
1.6
1.625
1.6153846153846154
1.619047619047619
1.6176470588235294
1.6181818181818182
1.6179775280898876
1.6180555555555556
1.6180257510729614
1.6180371352785146

Calculation of PHI from limit of Fibonacci divisions

If we suppose that the ratio of consecutive terms in the Fibonacci sequence do approach a limit we can use this to find the value of phi.

IF we denote the nth Fibonacci number by f_n and the nth ratio as r_nthen

r_n = f_n+1 / f_n

But using the definition of a Fibonacci number:

f_n+1 = f_n + f_n-1

then the ratio is

r_n = (f_n + f_n-1 )
f_n

Which can be simplified to be

r_n = 1 + f_n-1/f_n

but f_n-1/ f = 1/r_n-1

so

r_n = 1 + 1/r_n-1

Now supposing that the ratios tend to a limit p as n tends to infinity then p is the solution of the equation

p = 1 + 1/p

which can be re-arranged to give

p² – p – 1 = 0

and can be solved using the quadratic equation to give

p = (1 + √5)/2
and
p = (1 – √5)/2

The first of these solutions happens to be the golden ratio PHI or Φ

If anyone has a proof that these ratios do infact approach a limit please include it in the comments of this post.

Also see

• Arithmetic Series and Progression (2)

Note//In more advanced maths a^b is often taken to mean exp(b ln(a))

There are a number of rules regarding how to manipulate indices of which the most important are listed below:

1. a^b x a^c = a^b+c
   since we have axa b times time axa c times giving axa b+c times
2. a^b ÷ a^c = a^b-c
   by similar logic to point 1
3. (a^b)^c = a^bc
   since (a^b)^c = a^b x a^b… c times…a^b
   but by (1) we get a^b+b+…+b = a^bc
4. a^1/b = ^b√a
   Since by (3) (a^1/b)^b = a^b/b = a
   but re arranging we get the result
   a^1/b = ^b√a

If there any other rules that I haven’t included and aren’t immediately obvious from the above rules please leave them in the comments below

By David Woodford

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However not all fraction have the same denominator but we may still need to all them. To do this we make use of the fact that if you multiply the top and bottom of a fraction by the same number then it is still of the same value
for example
3/4 = (3 x 2) / (4 x 2) = 6/8

We can use this to add fractions as we can multiply the tops and bottoms of a fraction so that both the fractions we are adding have the same denominator (often called the lowest common denominator of LCD). This is can often be found by multiplying the denominators together.

eg)
3/4 + 2/3 = (3 x 3)/ (4 x 3) + (2 x 4)/(3 x 4) = 9/12 + 8/12

but in some examples you can then cancel these new fractions down to get simpler fractions with the same denominator

eg)
5/6 + 3/8 = (5×8)/(6×8) + (3×6)/(8×6) = 40/48 + 18/48
but both of these fractions can be cancelled down by dividing by 2 whilst keeping the denominator the same to get
20/24 + 9/24 = 29/24

Also see

• Matrix C++ program (0)

In the example i have written the series as
2 + 5 + 8 + 11 + 14 + 17 + …
instead of
2 , 5 , 8 , 11 , 14 , 17, …
because it is often interesting to look at sum of n items in a series. The sum S_n of the first n terms in a series can be found using
S_n = n(2a+(n-1)d)/2

Proof of sum of series
We can show that
S_n = n(2a+(n-1)d)/2
as follows

Write out the series in order and then in reverse order as follows
S_n = a + (a+d) + (a+2d) + …. + (a-(n-1)d) + (a+nd)
S_n = (a+(n-1)d) + (a+(n-2)d) + (a+(n-3)d) + …. + (a+d) + a
Now if we add each of the terms in the two sums we get
2S_n = (2a+(n-1)d) + (2a+(n-1)d) + (2a+(n-1)d) + … + (2a+(n-1)d) + (2a+(n-1)d)

where the 2a+(n-1)d repeatsn times. This is because we adding the rth term with the(n-r)th term hence we doing
(a+(r-1)d)+(a+(n-r)d)=2a+(n-1)d
so
2S_n=n(2a+(n-1)d)
therefore
S_n=n(2a+(n-1)d)/2

Also see

• How to use Surds – Add, Multiply and Rationalise (1)
• Geometric Series/Sum (0)
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When taking a random sample it is much better to use a confidence interval for your results rather than just giving the mean, because it gives an idea of how reliable your mean is. This is because you can calculate an average value from a set of completely random results but this doesn’t mean that you can have any certainty that the next result will be similar to the mean (since we have stated that the results are completely random they are no more likely to be close to the mean than any other value).

Calculating a confidence interval (for a normal distribution)

When calculating a confidence interval you must first decide on the percentage certainty that you are going to use for the interval (a common value to use is a 95% confidence interval)

You then use the reverse tables for the normal distribution to work the value of the standardised normally distributed variable to use.
Note: You must use the value half way between the certainty level and 100%, ie if you want a 95% confidence interval use 97.5% since you only want 2.5% on either side of the distribution.

You can now calculate the interval. To do this you need the standard deviation and mean of the sample. However to correct the standard deviation for the entire sample of possible tests divide by the square root of the number of items in your sample

ie) if you sample of n items is Y and the entire sample of possible results is X
sd(X) = sd(Y)/√n

Now the confidence interval is

X – z sd(X), X + z sd(X)

Where z is the value obtained from the inverse tables.

Also see

• The new Trevor Pythagoras Book Store (0)
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