The Student Room Group
Reply 1
Pure 5 is part of the normal A-level course and as such can be done with nothing more than P1, P2 and P4 under your belt. P6 is part of the further course and you need to understand a fair bit of P4/P5 to do well in it. Go for P5
Reply 2
thanks bekaboo, my maths teachers say that P6 is a new branch of mathematics alltogether, whereas P5 leads on (advances on) P1, p2 ... and so, for my further maths a-level i have chosen D2, M3 and (P5/P6). but i am attracted to p6, because it does not have integration and i heard it has lower grade boundaries. my teachers also say that P6 is easier, how far do you think this is true? anyone?
Reply 3
It depends on what exam board you're with. In my case (Edexcel), I felt that P4 and P6 shared more in common than P5, and the stuff in P6 (like matrices and induction) was more useful to me.
Reply 4
yup, sorry i forgot to mention i am with the EDEXCEL exam board. thanks for the replys so far! :smile:
Reply 5
KingAS
i need to chose between Pure maths 5 and P6. some say that p6 is somewhat easier.

for those of ya who are doin it, what do ya think?

I loved them both but I love P6 way more. I think P6 is quite a bit more difficult though than P5. Do you know what topics are in the P5/6 syallbuses on your exam board? Coz we might be able to tell you which are good topics and which aren't!
Reply 6
I hated P5 and I found P6 infinitely easier (although to be honest I did not get a much higher mark). P6 introduces completely new concepts but P5 is based on P1-4 except much harder. I found P6 a lot more interesting, especially matrices :smile:
Reply 7
the specifications are pretty long, but here is a general outline (main topic areas):
P5
Coordinate Systems
Hyperbolic functions
Differentiation
Integration

P6
Complex Numbers
Matrix Algebra
Vectors
Maclaurin and Taylor series
Numerical Methods
Reply 8
i guess that doesnt mean too much, but thanks for the suggestions, i think i will do P6 because most of my maths class are as well. Any ideas / thoughts about the topics above?

BUMP
Reply 9
Bump
Reply 10
u gotta ask urself if ur more of a calculus guy or an algebra guy.

i find algebra easier, hence i found p6 extrememly easier than p5. ide recommend do both if ur considering a degree in mathematics
Reply 11
yup i think i am pretty much the same... i am thinking of doing mech engineering
Reply 12
i doing both, but i don't like 3d vectors, i just can't get my head around them... if i had to choose, and bear in mind i am not edexcel, i would pick p5... it just depedns on which topics you like, but like euclid said, consider both if you are doign a maths degree...
Reply 13
anyone know what the Maclaurin and Taylor series is about??
it sounds pretty complicated, how hard is it?
Reply 14
Taylor series invovle writing out mathematical functions as polynomails, roughly of the form:
f(x) = a0+a1x+a2x²+..anx^n
The resulting polynomial can be used to approximate the value of the function at a certain x. For example, for finding an approximation of e^x by a polynomial, let's consider:
f(x) = e^x
f(x) = a0+a1x, which is a linear polynomial (has straight line graph). A good idea would be to use the tangent to the curve of e^x at (0, 1).
We know that f'(x)=f(x)=e^x, so at x=0, f'(0)=e^0=1. That is
f(x) = 1x+1 (since the y-intercept is 1)

We now know that e^x is approximately x+1. But this isn't really accurate, so we take it a little further.
f(x) = e^x
g(x) = a0+a1x+a2x², which is a second degree polynomial.
Now, instead of using a line to approximate g(x), let's use a curve: a parabola. The idea is to keep close to the curve of e^x as long as possible. To find a0, a1, a2, we must have f(x)=g(x), f'(x)=g'(x) and f''(x)=g''(x) at x=0.
f(x)=e^x, f'(x)=e^x, f''(x)=e^x
g(x)=a0+a1x+a2x², g'(x)=a1+2a2x, g''(x)=2a2
Using x=0 (i.e. f(0)=1, f'(0)=1, f''(0)=1, g(0)=a0, g'(0)=a1 and g''(0)=2a2) and solving f(x)=g(x), f'(x)=g'(x) and f''(x)=g''(x), we'll find that a0=1, a1=1 and a2=1/2.
So: g(x)=1+x+x²/2

e^x =~ 1+x+x²/2, which is a better approximation.

We can also take this a step further, taking third degree polynomials and third derivatives and following the usual method to find an even better approximation. So on and so forth. Sometimes, like with the case of e^x, you can spot a pattern:
e^x =~ 1 + x + x^2/2 + x^3/6 + ...
i.e.
e^x =~ 1 + x + x^2/2! + x^3/3! + x^4/4! + ..., which is known as a power series.

Generally speaking, Taylor series are used for finding (good) approximations for f(x) for values of x near a by using a polynomial. We used a=0 to approximate e^x. To obtain a suitable polynomial that behaves like f(x) for x near a, we need its value for x=a and the value for its first n-derivatives to match the values for f(x) and its derivatives all at x=a. That is:
f(a)=g(a)
f'(a)=g'(a)
f''(a)=g''(a)
...
f'^(n)(a)=g'^(n)(a)

Let's assume g(x)=b0+b1(x-a)+b2(x-2)²+b3(x-3)³+...bn(x-1)^n, which is the most suitable polynomial form that we can use. Then by differentiation:
g'(x) = b1+2b2(x-a)+3b3(x-a)²+...+nbn(x-a)^(n-1)
g''(x)=2b2+3(2)b3(x-a)+...+n(n-1)bn(x-a)^(n-2)
...
g'^(n)=n(n-1)(n-2)...(3)(2)(bn)=(bn)n!

We must also have:
f(a)=g(a)=b0 => b0=f(a)
f'(a)=g'(a)=b1 => b1=f'(a)
f''(a)=g''(a)=2b2 => b2=f''(a)/2
[this]
...
f'^(n)(a)=g'^(n)(a)=(bn)n! => bn=f'^(n)(a)/n!

By substituition, we get the forumla:
f(x) =~ f(a) + f'(a)(x-a) + f''(a)/2! . (x-a)² + ... f'^(n)(a)/n! . (x-a)^n
This is known as Taylor's approximation (Taylor series!).
If a=0:
f(x) =~ f(0) + f'(0)x + f''(0)/2! . + ... f'^(n)(0)/n! . x^n
This is known as Maclaurin's approximation (Maclaurin series!).

That's the jist of it.

btw, f'^(n)(x) is meant to be the nth derivative of f(x).
Reply 15
do both of them. If you are doing NatSci in uni, you'll need to know all there is to know in p5 and p6.
Do both!