## Monday, 30 January 2012

### Feedback Post

What have you liked about the blog so far ?
Any feedback would be extremely helpful.

I'm very busy with work, so I don't get to update it as frequently as I wish. I'm also trying to keep the content general as possible. What i will do within the next couple of days, I will label each post according to which specification and module it is in.

e.g. Transformation of Functions, at the top of the post i will list the module, the topic is included in the exam board specification.

Any feedback, or requests to be made for a post on a specific topic, please comment on this post or email on :

mathematicsthesolver@gmail.com

## Tuesday, 24 January 2012

### The Binomial Expansion

• Edexcel, Module - C2, Chapter- Sequences and Series
• AQA, Module - C2, Chapter - Sequences and Series
• OCR, Module - C2, Chapter - Sequences and Series
The Sequence and Series chapter in c2, is quite big, so I will divide it into 3 / 4 posts. a) Binomial Expansion 1 b) Binomial Expansion 2, c) Geometric Sequences 1 ,d) Geometric Sequences 2.

The Binomial Expansion, is a theorem which allows us to expand (a + b)^n, where n is an integer. (a and b are just what i've used it, they can be any letters).

Why we need it ?

Let's say we have to expand the following terms :

(x + y)^1 = x + y
(x + y)^2 = x^2 + 2xy + y^2
(x + y)^3 = (x + y)(x + y)^2
= (x + y)(x^2 + 2xy + y^2)
= x^3 + 3x^2y + 3xy^2 + y^3

As we see, when increasing the power, (n) we get more and more terms, and it becomes more and more confusing, say we had to expand (x + y)^7, this would mean we would get 8 terms, after expanding the bracket 7 times, and then simplifying it. This would be a tedious and extremely long !

The Binomial theorem therefore allows us to expand the binomials ( two terms e.g. x + y or x + 3y or a - 3b). In each term of the expansion, there will be a coefficient of the term, these coefficients are either determined by Pascal's Triangle or Factorial Formula.

Let's see how this works, with the examples above... Let's expand (x + y)^3 using the theorem :

Binomial Theorem = To expand the binomial (x + y)^n, it can be written in the form :

 This is the binomial coefficient, where n is the power and k is the increasing value till we reach n.

Let's use take our example using the theorem :

(x + y)^2 = (2,0) x^2 y^0 + (2,1) x^1 y^1 + (2,2) x^0 y^2

*The brackets are the binomial coefficients in blue , which are calculated using the formula :

* Notice the x terms start with the power n (in this case 2), and decrease by one power the next term, till the reach the power 0, simultaneously the y terms start with the power 0, and increase by one power the next term, till they reach the power n.

* We finish the binomial expansion when the binomial coefficient is (n,n) in this case (2,2), and when the x term has reached to the power 0 (which is 1), and the y term is to power n (in this case 2) . (Remember anything to the power 0 = 1 and anything to the power 1 = itself )

Let's simplify it down further, these are the binomial coefficients for each term :

! = Factorial (it calculates the product of all the positive integers less than or equal to some integer n)
(e.g. 4! = 4 x 3 x 2 x 1 = 24)

Note :  1! = 1 and  0! = 1

(2,0) = 2! / [ 0! (2 -0)! ]
= 2 / [ 1 x 2 ]
= 2 / 2
= 1

(2,1) = 2! / [ 1! (2 -1)! ]

= 2 / [ 1 x 1 ]
= 2 /1
= 2

(2,2) = 2! / [ 2! (2 -2)! ]

= 2 / [ 2 x 1]
= 2 / 2
=1

So we have calculated the binomial coefficients for the terms, giving us :

(x + y)^2 = x^2 + 2xy + y^2

Pascal's Triangle ( Second way to find binomial coefficients - helps when n is small)

Another way to find the binomial cofficients, is the Pascal's Triangle. We start with n = 0, and start with number 1. Each row, will list the binomial cofficients in order of the expansion, each starting with 1 and ending with 1, depending on n.

To find the coefficient underneath, two gaps, we simply add the coefficients above it.

n = 0                                                                              1
n = 1                                                                  1                     1
n = 2                                                      1                       2                   1
n = 3                                         1                       3                      3                     1
n = 4                            1                        4                       6                   4                         1
n = 5               1                        5                      10                    10                    5                          1

* The Triangle goes on as n increases, and the number of coefficients is always ( n + 1)
* The Triangle is symmetrical halfway, so we see that the first half of the coefficients of terms, is always the same as the second half
* Each row of the triangle starts and ends with 1 (meaning the first coefficient and last coefficient of any expansion is always 1)

Double check the expansion (x + y)^2 , where n = 2,
Check the row n = 2

The binomial coefficients should be 1, 2 and 1

1x^2 + 2xy + 1y^2

So we've seen how to expand expressions in the form (x + y)^n, and use two methods of working out the binomial coefficients. In the next post I will show you a couple of examples, and another expansion theorem, which allows us to expand algeabraic expressions in the form (1 + x)^ n.

## Saturday, 21 January 2012

### Equation of a Circle

•  Edexcel, Module - C2, Chapter - Coordinate Geometry
• AQA, Module - C1, Chapter - Coordinate Geometry
• OCR, Module - C1, Chapter - Coordinate Geometry

In C1, part of the Coordinate Geometry Chapter, we learnt about linear equations and graphing them. In C2, we learn about the equations of circles, from which we can graph them, aswell as solving problems again involving tangents. Remember in GCSE, we learn about Circle Theorems, here are a few we need to know, which will come in use in this chapter :

Circle Terminology !

Circumference - The distance around a circle, given by the formula 2pi * r.

Diameter - The distance from one point on the circumference, to another, passing through the centre of the circle. It is twice the radius (2r)

Radius - A line which is from the centre of the circle, to any point of on the circumference of the circle. It is half the diameter.

Area of a circle is given by the formula - pi * (r^2)

Circle Theorems Recap !

1) The angle in a semicircle is a right angle (An angle formed by drawing lines from the ends of a diameter to its circumference, will form a right angle)

2) The perpendicular from the centre of the chord, bisects the chord.
(A chord is a line that has both endpoints on the circumference on the circle.)

3)The perpendicularity of the radius and tangent.
(A tangent to the circle, forms a right angle, with the circle's radius, at the point of contact with a tagnent).

 Showing the 3rd theorem, which shows a line from the centre of the circle till the circumference (Radius), when it meets with a tangent to the circle at a point, it will form a right angle !

Equation of a Circle

The equation of a circle is given in the form : (x-a)2 + (y-b)2 = r

a = x -coordinate of centre
b= y-coordinate of centre

e.g.1 ) What is the equation of a circle when the centre of the circle is (-4,5) and the diameter is 10.

centre of circle (-4,5), so a = -4  and b = 5
radius = half the diameter so 10/2 = 5
so   (x + 4)2 + (y - 5)2 = 5

Just leave it in the form 5 squared, rather than writing 25.

e.g.2) What is the centre of the circle, and radius, given this equation of the circle :

(x)2 + (y+8)2 = 49

Centre of Circle -a= 0, and b = 8. So centre of circle is (0, -8 )
Radius is the square root of 49, some people still write + / - 7, even knowing a length can't be negative. so the radius is 7.

We've met our first point, knowing how to derive the equation of a circle, given the centre points and the radius, and vice versa.

Finding the centre of a circle, through Completing the Square

If we are given the equation of the circle to be :

x2 + 6x + y2 - 8y - 11 = 0 , and we have to determine the centre of the circle and radius. Here, we can see it's not in the usual form, but still it is an equation of a circle. We just have to complete the square (look under the c1 completing the square post if you've forgotten) :

complete the square for x terms and y terms seperately:

so  x2 + 6x  becomes = (x+3)2 - 9
so y2 - 8y becomes = (y - 4)2 - 16

we move the -11 to the other side, so we get :

(x+3)2 - 9 + (y - 4)2 - 16 = 11

(x+3)2 + (y - 4)2 - 25 = 11

(x+3)2 - 9 + (y - 4)2 = 36

(x+3)2 + (y - 4)2 = 62

Now, we have to the equation of the cirlce in the desired form :
(x- a)2 + (y - b)2 = r2

The centre of the circle is (-3,4) and the radius is 6.

This is all to the coordinate geometry chapter, in terms of what we need to know. In the next post, I will post some questions and their model solutions that are quite common, which require some knowledge of GCSE theorems!

## Wednesday, 18 January 2012

### Algebraic Division,Factor Theorem and Remainder Theorem

•  Edexcel, Module - C2, Chapter - Algebra and Functions
• AQA, Module - C1, Chapter - Algebra
• OCR, Module - C2, Chapter - Algebra

First chapter of C2 is Algebra and Functions again. We firstly start off with Algebraic Division, this is basically the long division we've done with numbers, except now we have polynomials. Before proceeding, it's good to know what a polynomial is.. and how to factorise certain polynomials (cubics, quadratics).. if you've forgotten. look into my c1 posts.

We are required to divide polynomials by (x-a ) or (x+ a), where a is some integer. This division process can either result in (a being a factor, which will mean after the division is complete, the remainder will be 0, and we will obtain a quotient (which is usually a degree less of the polynomial being divided). OR the division can result in a remainder, in which a is not a factor.

I will show you the actual process and examples in my image notes, it is hard to write it on here. The process is always the same. A couple of common errors :

* Say we have x^3 + 3x + 2. and we have to divide it by some factor (x-a).. some students may not be able to continue after some point, it is simply because we have to include the x^2 term in the division, so we rewrite the polynomial as :  x^3 + 0x^2 + 3x + 2 divided by (x-a).  [Always include terms that are not present by writing 0x^a (a being whatever power).. this will line up the columns]

General Teminology

Factor - Something that is completely divisible by another thing, (no remainder) x+4 is a factor of x^2 + 8x + 16, 2 is a factor of 10, 10 is a factor of 150.. etc.

Remainder - The value obtained / left over when one polynomial is divided by another, where it is not a factor. The Remainder Theorem can be used to find the remainder, instead of long division.

Divisor - The polynomial you are dividing by. e.g x^2 - x + 1 divided by (x - 2 < divisor)

Quotient - The expression obtained as a result of dividing polynomials. e.g. ( in numbers 8 /4 = 2 - quotient)

Quotient * Divisor = Dividend (original expression)
or
(Quotient*Divisor) + Remainder = Dividend (original expression)

Factor Theorem

The Factor Theorem states, that if f(a)=0, then (x-a) is a factor. This is another way of checking if some expression is completely divisible by another.

Remainder Theorem
Same as the factor theorem, when you divide a polynomial by (x-a), the remainder will be f(a).

 Notes of Algebraic Division 2
 Notes for Algebraic Divsion 1

## Thursday, 5 January 2012

### C2 - Introduction

C2 is the second module, of the Alevel Maths which is core. It is in my opinion the hardest module of AS, for Edexcel, it is one of the longer modules of the whole course.The topics are :

Edexcel
1. Algebra and Functions - Algebraic Division, Factor Theorem, Remainder Theorem.
2. Coordinate Geometry - Equation of Circle, Use of certain circle properties
3. Sequences and Series - Geometric Series {Sum of a g. series, Sum to infinity, Proof of sum formula}, Binomial Expansion for (1+x)^n and (a+b) ^n.
4. Trigonometry - Sine and Cosine Rule, Area of Triangle, Radians, Arc length, area of a sector and segment, Sin,Cos and Tan Graphs, Use of Trig. Identities, Solving simple equations.
5. Exponentials and Logarithms - y=a* and the graph, Laws of Logarithms and Change of base formula.
6. Differentiation - Maxima and Minima problems, Stationery points, Increasing and Decreasing Functions.
7. Integration - Definite Integrals, Area under a curve is a definite integral, Trapezium Rule, Area bounded between a curve and line.
AQA

1. Algebra and Functions - Laws of Indices, Simple transformations on graphs
2. Sequences and Series - Arithmetic series {nth term, Sum of a. series}, Geometric series {Sum of finite g. series, sum to infinity}, Binomial Expansion (1+x)^n.
3. Trigonometry - Sine and Cosine Rule, Area of Triangle, Radians, Arc length, area of a sector and segment, Sin,Cos and Tan Graphs, Use of Trig. Identities, Solving simple equations.
4.  Exponentials and Logarithms - y=a* and the graph, Laws of Logarithms and Change of base formula.
5. Differentiation - Differentiate functions of form x^n. where n is a rational number.
6.  Integration - Integration of functions of form x^n, Trapezium Rule
OCR

1. Algebra - Factor theorem, Remainder theorem, algebraic division, a* graph, laws of logarithms, change of base of logs, and solving log equations.
2. Sequences and Series - Sigma Notation, Arithmetic & Geometric Series, Sum of arithmetic and geometric series, sum to infinity, Binomial Expansion.
3. Trigonometry - Sine and Cosine Rule, Area of Triangle, Radians, Arc length, area of a sector and segment, Sin,Cos and Tan Graphs, Use of Trig. Identities, Solving simple equations.
4.  Integration - Indefinite Integration, Evaluation of definite integrals, Finding area of a region bounded by a curve and line,. Trapezium Rule.

There is alot of new content, and alot which is expanded from the topics learnt in C1. To Achieve a top grade, you should ensure you can apply everything used in C1, these modules are synoptic, and they can ask you a c1 question in c2, or ask a question which relates to c1 content.

## Tuesday, 3 January 2012

### Indefinite Integration

•  Edexcel - C1, Integration
• AQA - C1, Integration
• OCR - C2, Integration

We already discussed one of the branches of calculus. The other is Integration, it's general purpose is to find the area between some given intervals. It can be used to find the area bounded by a curve, you will learn this in C2. Right now, we need to know what Indefinite Integration is. It is simply the reverse of Differentiation, if dy/dx is the derivative of some function, we can use integration to obtain the function y. It's also called Anti differentiation. When will integrate some function, we will add a constant c. (The constant of Integration).

That image is the general notation used for Integration. When we integrate, we first draw the S kind of line, then write the function, and then dx. (for now anyway). This shows we are integrating. We need to know how to integrate functions in the form : x^n.

As I said Integration is the reverse of Differentiation. So when we differentiate we multiply the power by the coefficient of the function, and subtract the power by 1. When integrating, we firstly raise the power by 1, and divide by this new power.

a)

*We usually write c (instead of constant) [  + c ]

I will also post some further examples in my image notes. In my image notes, I will also show you how to derive a the equation of the curve, when given dy/dx and a point on the curve, these question are usually 5/6 easy marks.

With this we end C1 here. I shall start C2 next week... I have done all model solutions for the Edexcel C1 Papers (From Jan 2005 - June 2011), these are available on request. I will start working on the model solutions for Solomon Papers aswell.

### Differentiating Functions with Examples and Applications of Differentiation

I will use two examples from the specification itself, which you should be able to differentiate :

1) y= (2x+5)(x-1)
First we expand the brackets, and simplify :

2x^2 - 2x + 5x -5
y = 2x^2 + 3x -5, now we can differentiate so :
dy/dx = 4x + 3

I will show the 2nd example, in my set of image notes.

Application of Differentiation to Equations of Tangents

If we have a curve, and a tangent to some point on the curve. We can find the equation of the tangent. How ?

If we know the coordinates of that point. Firstly we differentiate the curve, and substitute the x-coordinate to get the gradient of the curve at the point. Say, if the tangent is perpendicular to the curve. We know that the gradient of this tangent, will be -1 divided by the gradient of the curve, we just got. Using the gradient, and the coordinates, we can use the y -y1 = m(x-x1) formula to find out the equation.

Example

This question is taken from the Edexcel C1 Jan 2006 Paper. (Question 9)