Red Fraction

Somewhere down the line during the course of this academic year, Statistics stopped becoming a nuisance, and started becoming a delight. Don’t ask me why; I never thought the day would arrive where I admitted to enjoying A-Level Maths lessons. Because, quite frankly, last year was torturous.

Previously, I couldn’t extract any real interest from the subject as a whole. My initial reasoning for taking Statistics at A-Level was because: I lacked the natural ability to cope in French (the A* at GCSE was not a fair reflection of my understanding in the subject; though I didn’t revise particularly hard for the exams, I just knew what the examiners were expecting of me), I despised all three Sciences, and morsels of information about Computing meant that I could never embrace the idea of learning programming (in retrospect, I wish I’d tried learning it outside of school).

I think half…or maybe eleven-fifteenths (HAHAHA! Maths joke! HAHAHA!) of the reason for my sudden reversal in exasperation at the subject is my comprehension of it. Yes, that must be it. Had you asked me a year ago to convert a random improper fraction like x² + 3x – 2/(x + 1)(x – 3) into a partial fraction, then I wouldn’t have had the slightest clue where to begin. Probably because it wasn’t on the syllabus at AS, but whatever *glances from side to side*. Now, however, I know how to do it with ease!

WARNING: The following contains a heavy use of jargon that may or may not be familiar to you. For your wellbeing (and the wellbeing of your PC monitor, as in protection from flying fists), please have patience in the reading of the following. In other words, don't just skip it without even looking!

x² + 3x – 2/(x + 1)(x – 3) expanded = x² + 3x – 2/x² – 2x – 3.

Using long division, we discover that x² + 3x – 2/x² – 2x – 3 = 1, with a remainder of 5x + 1.

Therefore x² + 3x – 2/(x + 1)(x – 3) = 1 + 5x + 1/(x + 1)(x – 3).

Now, to convert into a partial fraction, 5x + 1/(x + 1)(x – 3) must be expressed in the form A/(x + 1) + B/(x – 3), where A and B are constants to be found.

If 5x + 1/(x + 1)(x – 3) = A/(x + 1) + B/(x – 3), then…

5x + 1/(x + 1)(x – 3) = [A(x – 3) + B(x + 1)]/(x + 1)(x – 3)
(multiplying A/(x + 1) by (x + 1)(x – 3) gives A(x – 3) as (x +1)/(x + 1) = 1).

Denominators are now the same, so…

5x + 1 = A(x – 3) + B(x + 1)

If x = 3, then…

5(3) + 1 = A(3 – 3) + B(3 + 1)
15 + 1 = A(0) + B(4)
16 = 4B (A(0) = 0, so A disappears)
16/4 = B
4 = B

And if x = –1, then…

5(–1) + 1 = A(–1 – 3) + B(–1 + 1)
–5 + 1 = A(–4) + B(0)
–4 = –4A (B(0) = 0, so B disappears)
–4/–4 = A
1 = A

Alternatively, since we already had the value of B, we could have put it into the original equation (where x = 0) to work out A (We’ll use this as a test to see if the values of A agree with one another).

5(0) + 1 = A(0 – 3) + 4(0 + 1)
1 = A(–3) +4(1)
1 = –3A + 4
1 – 4 = –3A
–3 = –3A
–3/–3 = A
1 = A

As you can see, A does in fact equal 1, meaning we just stick the values of A and B back into the fractions above…

5x + 1/(x + 1)(x – 3) = A/(x + 1) + B/(x – 3) is now

5x + 1/(x + 1)(x – 3) = 1/(x + 1) + 4/(x – 3)

(note that 1/(x + 1) could also be written as (x + 1)‾¹ and 4/(x – 3) could be written as 4(x – 3) ‾¹, since 1/x = x‾¹)

That’s it! I'd like to convince more people that Maths is, indeed, a very pleasurable subject. For example, when you find out why sin θ cot θ sec θ = 1, it's brilliant!

Prove sin θ cot θ sec θ = 1?

Well, cot θ = cos θ/sin θ.
And sec θ = 1/cos θ.

Therefore, sin θ cot θ sec θ = sin θ(cos θ/sin θ)(1/cos θ).

sin θ(cos θ/sin θ) = cos θ, and cos θ(1/cos θ) = 1.

Meaning...sin θ cot θ sec θ = 1!

I love that! And honestly, none of this is irony!

If you can understand why I now find this sort of thing fun to do, well done!

Moreover, if you made it to this part of the article, and read the entire thing without groaning, or skipping the mathematical section, a warm appraisal I hand out to you.

Don’t worry; future articles will not be like this one *cheeky grin*!