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 axⁿ and 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)ⁿ 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.

Pascal's Triangle

Pascal's Triangle has it's roots far before Blaise Pascal arranged it in the trangular form we all know (and love?) today, but what actually is Pascal's Triangle, why's it useful and what are it's properties?

The first 13 rows of the triangle are listed above, but what is there that we can notice? Well each number is the sum of the two directly above it, and that is the key to constructing a Pascal Triangle of your own. Other than creating some very interesting and intriguing patterns, some of which I will explain...

Line one is the diagonal line of ones, line two is the counting numbers. The next line (the third line) is the triangle numbers. Here's where it gets even more fascinating, and quite frankly beautiful; if you shade the even numbers with one colour and the odd numbers with another colour you end up with Sierpinski's Triangle, which looks like...

And as you can see, it's actually quite a wonderful piece of art, surely this is the epitome of eloquence within Maths?

For the more Mathematical minded among us you may find it more eloquent that each row of the triangle adds to be 2^(n-1), n being the row number. Another use of Pascal's Triangle? Showing the coefficients in the binomial expansion. Say we choose (x+1)^4, if we look at Pascals Triangle we can see that the coefficients will be: 1 4 6 4 1. The actual expansion is; x^4+4x^3+6x^2+4x+1, can you also see a pattern here? But that's another post for another time...

This surely is the definition of beauty within Maths, how intricate a very simple piece of Maths can become.

I'd like to see what you can do now though, show me your best Pascal Triangles or Sierpinski Triangles in the comments or tag me in a post on my twitter: http://twitter.com/#!/EloquentMath