Those of you guys that are preparing for your A-Level exams or are getting ready to prepare for your advance Mathematical Science course at HE (Higher Education) purses. Here are the most important books you should buy and learn, not just for your exams but for your knowledge in science. It cover this areas of Maths, statistic, Mechanics and Discrete Maths.

Remember, your A-Level covers all these areas. I would want to use this opportunity to start a thread on advance Mathematics and Mechanics. I want to start from

1. Trigonometric
2. Differentiation
3. Integration and Limits
4. Power Series
5. Vectors
6. Complex Numbers
7. Taylor series
8. L' Hopital's Rule
9. Maclaurine Series

I am not good at all these topic, but have a very grounded knowledge on them. Anyone interested in this topic shoould contribute for our knowledge repository.

I personally will reference a lot from this book [url=http://www.amazon.com/gp/redirect.html?ie=UTF8&location=http%3A%2F%2Fwww.amazon.com%2Fs%3Fie%3DUTF8%26x%3D0%26ref%255F%3Dnb%255Fss%255Fgw%26y%3D0%26field-keywords%3DMEI%2520A2%2520Pure%2520Mathematics%253A%2520C3%2520-%2520C4%2520%2528MEI%2520Structured%2520Mathematics%2520%2528A%252BAS%2520Level%2529%2529%26url%3Dsearch-alias%253Daps&tag=wwwopen2blogc-20&linkCode=ur2&camp=1789&creative=390957"]MEI A2 Pure Mathematics: C3 - C4 (MEI Structured Mathematics (A+AS Level))[/url]

Admin, please can you check if there are any mod_plugin for Mathematic for us and install it. if it is not too much to ask. Thanks.

I recommend any Mathematics text or study guide written by MEI (Mathematic in Education and Industry). They are the best A-Level body in the UK.


Progression

After studying these topics, you would be ready for ANY HE (Higher Education). You can study

1. BSc Mathematics
2. BSc Applied Mathematics
3. BSc Financial Mathematics
4.BSc Computer Science
5. BSc Statistics with Economics

The word Trigonometry comes from the Greek words : Treis = three, Gonia = angle and Metron= measure. The Early Greeks developed the subject by studying the relationship between the arc of the circle - the measure of the central angle - and the chord of the arc.

Initially it was used in Astronomy but later it was much used in Architecture, Navigation, Surveying and Engineering, but in the last two centuries it has been used more for Mathematical Analysis and for repeating Waves and Periodic Phenomena

Trigonometric is straight forward but if you are struggling with it let use know. Remember you should be able to prove that

(sin)^2theta + (co s)^2theta = 1

theta = The greak symbol teta
^ = square symbol

x^2 + Y^2 = r^2


=> (x^2/r^2 ) + (y^2/r^2) = 1             divide by r^2

=> but  x/r  = (co s)theta y/r = (sin)theta     by definition

.
,  (co s)theat + (sin)theta = 1


Remember x/r  = SOHCAHTOA = (co s) theat = Adjacent/HYP

Remember y/r  = SOHCAHTOA = (sin) theat = OPP/HYP

Remember (x, y)
x = horizontal
y = vertical

Differentiation

What is differentiation? In calculus, a branch of mathematics, the derivative is a measurement of how a function changes when the values of its inputs change.

In GCSE we delt with tangent of a point on a curve. To work out the gradient of a tangent we did

         y2 - y1
m =  ----------
          x2  x1

Two point must be given i.e. (2, 3), (2, 4) of a equation for example y = 3x + 1/2 taking th range -3 < x < 4

What is a gradient? It is how stip a line is at any point from point A to B. i.e.

The above graph is a function of y = 3x + 1/2 taken from a range of -3 < x < 4

If you take a tangent at point (2, 4) you can work out the slop? The small change in y divided by small chnage in x

0 - 4 4
m = -------- = -- = 2
0 - 2 2

The slop at cordinate (2, 4) is 2

You can take two points on a curve and find the slop i.e.

(2, 4) and (6, 10)


10 - 4 6 3
m = ------------ = ------ = ----
6 - 2 4 2

But how about if you want an instanteous change? Like

(2, 4) and you want (2.1, 4), (2.2, 4), (2.3, 4) ,  (n-1, 4)?


It is very tedious the previous way and besides you are given a cordinate points. If cordinate points are not given just the function like y = e^2x, find the instanteous change for point A to B

First, if we need to find the gradient function of y= e^2x, given any tangent point on a curve I will quickly get the gradient by substituting the values of x.

The process of finding the gradient function of a function is called differentiation. There are more complex function that you can never find using normal O-Level method to find i.e

y = mx + c -------> O-Level


Ok, let do some small reading.

There three main rules for differentiation

1. Chain Rule
2. Product Rule
3. Quotient Rule

There is this one called partial differentiation. This is when x and y are mixed up in a function. Remember, you differentiate to a respect of a variable.

i.e. let say

y = X^2

You want to find y when you know x. If x=2, them y=4

But, if you have y = x^2 +y, it becomes complicated. You cannot find y even if x was give. This sort of problem we use partial differentiation to get the value of y.

When you differentiate we say

f(x) = X^3


We say we want the quantity of y with the respect to small changes in x. Hence, d mean delta, which means small change

dy
---- = 3x^2
dx


What just happened?

            kn^(n-1)

K = constant i.e. 3
n = unknown i.e. x

dy
----    = 3x^(3-1)  = 3x^2
dx



Now solve

1. y= 2x^2 +3x find   dy
                            ----
                              dx

2. Find dy
------ for each of the following:
dx

(a) x=3t^2, y=2t at t=2
(b) x=sin^2 theta, y=co s theta . sin theta at theta= pi/6
(c) x=2t^3, y=2/t^2 at t=2


A-Level Question.

Find the gradient at the point where t = 6 on the curve given by

y = t^3 - 10t         and      x = t^2 + 10


Below is the graph the above two functions

For those that are very much advance in calculus, here is what you can do for me to keep you busy while I bring them up to speed.

Use Leibniz formula, given below, for the repeated differentiation of a product, to find


     d^6y
   ---------  for x^3e^(2x)
     dx^6


  D^ny = D^n(uv) =uD^nv+ nC1DuD^n-1v + nC2D^2uD^n-2v + nC3D^3uD^n-3v + ,

Below is how this equation looks

If u and v are two functions of x, then the derivative of the product uv is given by

Picture below

Example if we have function y such that x are two functions of y

u =  (2x^2 + 6x)

and

v = (2x^3 + 5x^2)

This is called the product rule.

  d[i]u[/i]
----  = 4x + 6
  d[i]x[/i]



  d[i]v[/i]
----  = 6x^2 + 10x
  d[i]x[/i]


  d[i]y[/i]
  ----   = (2x^2 + 6x)(6x^2+10x) + (2x^3 + 5x^2)(4x + 6)
  d[i]x[/i]


  After Factorization, I got

  dy
  ---  = 20x^4 + 88x^3 + 90x^2
  dx

Another Rule is the quotient rule

Pls could you give a detailed lecture on the applications of differentiation in terms of change in volume, change in area, etc as well as an introduction to the applications of integral calculus. Thanks in anticipation of your positive response.

@Godisking, it looks like you are advanced in differential calculus. What you are after is in the application of calculus hence, Applied Mathematics. We are still learning to differentiate simple equation and understanding the reason why we do it.

However, if you are into differential application, here is what I want you to do for me. If you get it right, then I will open a new thread on Differentail Application


A close cylindrical tank of radius 50ft and height of 72ft is to be painted on the top and the sides with a paint of thickest 0.01in. About how many gallons of paint (231in^3 = 1 gallon) are required?

Hint: The volume of a cylinder is given V = pir^2h

To further help you. The close top cyclinder is like this

I am doing this for you now. Find the change in dv using the product rule. Replace the value with v, h and r like the one below

if you look at volume of a cyclinder the formula looks like a product rule. Two functions u and v are multiplied to give you V. i.e. pir^2 * h. Remember, that pi is a constant variable. It does not change. Hence, it is like 2x. 2 here is a constant and will never change. So if you use the product rule to find the small changes as v gets closer to 0.

I am expecting your answer.

If the above differential application is a bit difficult for now, perhaps you should try this and come back to it later on.

@Godisking, you ran away? Ok, it is easy to ask question but you have to put effort to defend your question or show eagerness to learn. Well, here is the solution for the question.

Don't misunderstand this thread or any other thread that I post. I am not here to lecture or teach anyone. I simple want to share knowledge with fellow Mathematicians.

I cannot begin to teach anyone the above topics from elementary. These are advanced topic for any Mechanical and Mathematical student or perhaps, some A-Level students. So far, I have been the only one contributing towards the thread just to encourage other minds alike to join. I do not claim to be expert in the subject but would want to develop an efriend with mind alike. So, below is the answer.

We now have

This is the solution for

I have another question for our readers brought to you by webmonkey. If you have any question please ask it here or type it in a word document and save it as a JPEG file and attach it here for our consideration

Here is another brain teaser for you over the weekend.

Some hints are given below

Here is another brainteaser for the weekend.

Note:
y and x are mixed up in this function. We cannot say y = f(x) or x = g(x), hence we need to use partial differentiation

See the solution here

I want to touch a little bit on complex numbers. What is complex number? In mathematics a number of the form a+bi where a and b are real numbers and i the imaginary number or is the square root of -1

It is the expression of the form

z=a+bi where a and b are real numbers and i= root of -1. The number a is called the real part, denoted Re(z) and b is called the imaginary part, denoted I'm(z). Note that i^2 = -1. You add, subtract, multiply and divide complex numbers in the manner that you would think from elementry algebra.

Here is the solution. Looks simple right?

Remember these argand formula for cartesian form and their polar form

Question to this solution

Solution continued,

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Vector Calculus

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