Friday, 16 March 2012


The idea of logarithms is that they are the reverse operation of exponents, that is the purpose of them. That is that if ab = x, then, logax = b. John Napier was the first person to introduce logarithms in the 1600s and they rapidly came into use, they were especially useful for one fact about logarithms.

That is that loga(xy) = loga(x) +  loga(y) . This made large multiplications a lot simpler and simply turned it into a problem of addition. And in fact we can prove this fact relatively simply, and I will do after proving one other property. That is that  loga(xc) = c loga(x).

Let loga(x) = b, if we multiply both sides by c and write loga(x) = b in exponent form we get:
c loga(x) = bc and ab = x. If we raise both sides of ab = x to the power of c we get:
abc =  xturning this into a log gives:
 loga(xc) = bc, and we know that bc = c loga(x), so this means that:
 loga(xc) = c loga(x), proving what we needed to know.

Now to prove that loga(xy) = loga(x) +  loga(y):
Let m = loga(x) and n = loga(y) and then w
rite these in exponent form 
x = aand y = athen we multiply these together to give:
× y = a× an = am+n, we can now take loga of both sides and evaluate 
loga(xy) = loga(a)m+n, then we use what we proved previously to get:
loga(xy)  = (m + n) logaa, note that logaa = 1 for all a.
loga(xy)  = m + n, we know from our first line that 
m = loga x and n = loga y so:
loga(xy) = loga(x) + loga(y), proving what we want!

We can utilise the last two properties to also prove another property (we can also use the last method to prove it too, but this is far easier).
Consider loga(x/y) and notice that this can be wrote as:
loga(x × y-1), which from using our last fact means that:
loga(x × y-1) = loga(x) + loga(y-1), using the first property (loga(xc) = c loga(x)) this means that:
loga(x/y) = loga(x) - loga(y), providing another identity.

And that really is it for the basics of logarithms, you can do a lot with them so make sure you can understand these simply properties, they will make other problems a lot, lot simpler.

Sunday, 11 March 2012

Introduction: Differential Equations

Differential Equations have an absolute massive range of uses and are one of the most useful applications from maths to the real world, they have massive influences in: quantum mechanics, electromagnetic equations, biology, economics and an absolute ton more in all of physics; read more about notable differential equations. First I need to explain what a differential equation exactly is. A differential equation is any equation that involves a derivative of any degree, examples of differential equations are:

Differential equations can have different orders, when only the first derivative is present it is a first order differential equation. When there is a second derivative is involved it is said to be a differential equation of the second order. Differential equations of higher orders follow the same pattern for what order it is.

Differential equations can be either linear or non-linear. A differential equation is said to be linear if the highest order derivative of the dependent variable y can be expressed as a linear function of y and the
lower order derivatives. Hence a second order differential equation is said to be linear it must be possible to express it in the form:

To solve a differential equation you wish to find y in terms of x. Some of these are very easy to solve and can be solved by algebra and calculus methods (analytical methods), and you may notice that the first of these examples is very easy to solve and can be solved with calculus. All we need to do is integrate and we will get the answer for all values of y as y = x+ c. But often it is not that easy to solve and you can not easily express y in terms of x, when this is the case numerical methods are very, very useful.

A numerical method is one that will find approximate solutions to the equation at a point you wish to find. There are a lot of numerical methods to solving differential equations, some more effective that others. I will introduce just the one of these in this post.

Euler's Method for Finding Solutions to Differential Equations
This method was created by the mathematician who seems to have influenced all areas of maths in some way or another, Leonard Euler. If we have some unknown function, but we have what its derivative equals and where one point is on the unknown function then we can find approximate y-values for a point further along the curve.

This method is most effective for small values of h. It can of course be repeated to give a more accurate value for y. If we know a point on some function is (3,5) and we want to find the point 3.4, rather than using a step size of 0.4, we could use one 0.1 and just repeated the algorithm 4 times. This will be more accurate because it will map the graph and it's gradient at each step of 0.1, rather than assuming the graph will have the same gradient the whole time.

This is just an introduction to what differential equations are and one relatively simple method of solving them. If you liked this please like us on Facebook.

Sunday, 4 March 2012

What is Anti-Matter?

To start with, antimatter really isn't that different from ordinary matter, it is composed of particles (anti-particles, but still particles nonetheless) just like normal matter is. In fact the only real difference between matter and antimatter is that (generally) they have the opposite (but exact same magnitude) charge to their matter counterparts.

Antimatter is, however, a relatively new concept for science. In fact the first real, serious suggestion that antimatter could even exist was in 1928, by the English theoretical physicist Paul Dirac. He started out with pretty modest intentions, he wanted to discover how an electron spins, something that was not known at the time. And he managed it, the original equation he came up with was a slightly messy looking one, but still rather neat for a problem that had alluded physicists for so long.

But Dirac wasn't satisfied there, he felt that the equation could be simplified further and he felt that notation was the key, he worked and eventually, he managed to create an equation to explain the spin of an electron with just a few terms.

He then decided that this was ready for full publication, but when he did he encountered a few problems. The solutions to the equation implied that if it was how the electron behaves it must have a particle that is identical to it, exact for it having opposite charge. It wasn't generally accepted that this could be true, but Dirac believed that the equation was constructed so elegantly and beautifully that it must be true, and in fact this particle was discovered to exist in 1932 by Carl Anderson, this was the positron.

The main useful property on anti-matter is that when it collides with matter, it annihilates. This means that all of the mass of the two particles is converted into energy, and it is a lot of energy. Using Einstein's famous equation, E = mc2, we can work out how much energy is produced when annihilating an electron and positron. They each have a mass of 9.11 × 10-31 kilograms and c is the speed of light, 299,792,458 metres per second. So the energy produced in the annihilation of an electron and positron is, E = 2 × 9.11 × 10-31 kg × 299,792,4582 m/s = 1.64 × 10-13 joules.

Now this may seem incredibly low, but this is for one pair of tiny particles. If we had a collection of electrons and positrons the mass of a grain of sand (about 2.3× 10-5 kilograms) this energy level jumps to 4.13 × 1012 joules. This is about 50 times more than the average energy a car uses in a year. From matter no greater than the size of a grain of sand.

The energy released is in the form of gamma radiation which although it can be harmful it has very useful properties too. It is only through annihilation that PET scans (positron emission tomography) are possible. Electrons is fired at where a scan is needed to be taken from, and with the aid of a radioactive isotope it collides with a positron and produces a gamma ray. The gamma radiation is picked up via detectors and the data that is received can be reconstructed to show a detailed 'slice' of the area.

Because of the massive amounts of energy that are produced from a relatively small mass it would seem reasonable to try and harness this energy to power, well, everything. But here lies the problem. It is very difficult to produce large amounts of antimatter artificially. CERN when fully operational can produce 10 million anti-matter particles a minute, but even at this rate it would take over 100 billion years to produce just 1 gram of anti-hydrogen.

As you might imagine it is also very difficult to store anti-matter, because when it interacts with matter it annihilates into energy. In fact anti-matter has only been stored for 16 minutes at most in the whole of history, this is yet another stumbling block as to why anti-matter is very hard to utilise.

Because of both of these factors it is estimated (by NASA in 1999) that one gram of anti-hydrogen would cost $62,500,000,000,000 (that is 62.5 trillion dollars!). Because of this massive sum of money for such small amounts of anti-matter, research into it is a big thing in modern physics.

Saturday, 3 March 2012

Proof of Differentiation

This post is essentially to prove to you why differentiation gives the gradient of a curve, this is known as differentiation from first principles. This does not require you to understand how differentiation works, this is a proof of the well known principle of differentiation and why it is true for all values. If you do want to read more on the basics of differentiation you may want to download my Core 1 revision guide.

To begin explaining why differentiation works we have to consider what the gradient of something actually is, and that is the rate of change of a graph. The rate of change is the: change in y/change in x (which is what dy/dx means). So if we have some function, f(x) and we want to find the gradient of it at any given point it may be sensible to think that to find the gradient of a curve at any point (x, f(x)) we do:

But, this means that the denominator is 0, so the gradient of the curve according to this is undefined, which makes no sense at all. So we need to think of another way to try and find the gradient of a curve.

Let us consider the graph y = 2x2 + 3x - 1, if we want to find the general gradient for any point x, let us consider the line created by the two x co-ordinates x and x + h, the y co-ordinates of these points are 2x2 + 3x - 1 and 2x2 + 4xh + 3x + 2h2 + 3h - 1 respectively. So the gradient of this line is:

So this says that the gradient of y = 2x2 + 3x - 1 is 4x + 3, which agrees with differentiation as we know it. But this is just an example that differentiation works, not a proof that it is true for all values. But, applying the same principles we used in our example we can prove it for all values.
Let f(x) = axn, consider the line from connecting two points x and x + h, the y co-ordinates for these points is axand a(x+h)n, the expansion for this may not seem immediately apparent but if you have encountered Pascal's triangle and the binomial expansion you will understand that a(x+h)can be expanded to an extent. To fully understand what I will be next doing you may want to read up on the binomial expansion. If you really do not understand this next step, leave a comment and I will explain to the best of my ability.

And that is it! Hopefully you managed to understand what I have done, but if any points have you confused, please leave a comment. If you liked this, why not like us on Facebook? Go on, you know you want to.