I did the C1, C2, C3 and M1 modules in Year 11, and C4, S1-4, M2-5, FP1-3 and D1-2 modules in Year 12, which when cashed in should yield 3A* grades but they are withholding the cashing in because they want me to consider doing re-sits for modules I didn't do too well in (like D2).

Is anyone else doing FM?

]]>luca-deltodesco wrote:: How are you finding FP1?

tiss easy, i already know all of it and more which probably helps

how comes you already know must of it?

can u give me a hand on series then please? i dont get a word they are tralkin about!

It would be easier for us if you posted specific questions.

]]>: How are you finding FP1?

tiss easy, i already know all of it and more which probably helps

how comes you already know must of it?

can u give me a hand on series then please? i dont get a word they are tralkin about!

]]>tiss easy, i already know all of it and more which probably helps

]]>I'm not really sure what I want to study yet.. but I was thinking about Warwick.. and Nottingham. I might do physics with philosophy - looks really interesting.

]]>so far ive done C1-C4, M1, M2, S1 and in january im taking FP1 and D1

Haha, I took one look at that D1 book and thought there's no way i'm ploughing through this lot! How are you finding FP1?

Over the summer i've done C4, FP1 and FP2, and i've almost finished M3 now. I'm gonna sit all these in january so i'll only have 2 to do in the summer, plus any resits if necessary. I've actually found M3 to be quite good, really interesting topics (well i find them interesting anyway) on centres of mass of solids and stuff. It seems about the same difficulty as M2 to me, just with different material. I think i'm gonna give M4 and maybe even M5 a go - the books are nice and slim From what i've done so far of the further maths material, the integration in FP2 is the fcuker - hopefully the questions in the exam papers won't be as tough as some of those in the book.

Wow there are lots of people doing further maths in your schools!! Where i'm at, there are only 5 people in my regular A2 math class, and one of them doesn't show up much!

I'm doing physics and ICT with the maths. Wanna do a theoretical physics degree at uni.. maybe Warwick or Imperial with any luck. I'm off to the Warwick open day tomorrow actually You two decided what you're gonna study and where yet?

]]>Actually we have quite a few taking further maths... 13 including me in my year, and about 7 in the year above.

What other subjects are you doing? Are you going to / have you applied for uni?

]]>so far ive done C1-C4, M1, M2, S1 and in january im taking FP1 and D1

in my school there are 4 people taking further maths (excluding me) and they're all the people i wouldn't have thought would take it, i would have thought the sort of 'intellectual' type people would, but its been the opposite, the people in my class who seemed to care the least when it came to academics have all done further maths ^^

]]>Hello.

I'm doing further maths as well, but i am doing it at school. I'm in year 12 though.. so I'm on the normal A level at the moment. How are you finding it so far? Are you in year 12 or 13?

Heya. That's cool, i'm in year 13. So which modules are you working on right now / doing over the rest of this year?

I'm not finding it too difficult so far. Did C1-C3 + M1, M2 and S1 last year and it was all fairly staright-forward. M2 was easily the most difficult of the lot for me, but I was pleasantly surprised by my grade when it came through My worst mark came in S1 simply cos it's far too boring to study!

Are there many other ppl taking it at your school?

]]>Numerical methods is basically just number crunching, the kind of tedious, repititious process that we normally leave to computers.

Well, that doesn't seem so bad. Maybe a bit useless, but not so difficult to learn.

]]>It's used to approximate values that can't be found analytically (square roots, for example).

(i) Interval bisection

This is probably the most boring of the three, because it is the slowest to get you to a certain accuracy. It involves making an initial interval that you know the answer is in, and then halving it repeatedly.

For example, let's try to find :sqrt2. The equation in this case would be x²-2=0.

The first interval is sometimes given to you, but if you have to find your own, you're looking for two numbers a and b such that f(a)<0 and f(b)>0.

A good first interval in this case would be (1,2).

1²-2 = -1 and 2²-2 = 2, so that works as an interval.

Now we find the halfway point [(a+b)/2] and put that into our function.

1.5²-2 = 0.25. This is on the "f(2) side" of 0, and so it replaces 2.

The new interval is (1,1.5).

Now do that lots of times until your interval is small enough.

f(1.25) = -0.4375 --> (1.25,1.5)

f(1.375) = -0.109375 --> (1.375,1.5)

f(1.4375) = 0.06640625 --> (1.375,1.4375)

...

Note that it's important to keep the as much of the number as you can. Rounding will cause inaccuracies in later stages. You'll have to keep going until one of two things happen. The question will either say to keep going until your interval is a certain size, or until you can state the answer to a certain accuracy.

That last interval would be enough for you to say that :sqrt2 ≈ 1.4, because both values give that when rounded to 1dp.

(ii) Linear interpolation

Similar to interval bisection, but a bit cleverer, harder and quicker. In the above example, we initially had that f(1) = -1 and f(2) = 2. As f(1) is closer to 0, you might want to guess the next value as, maybe, 1.3 instead of just blindly halving the interval all the time, and that's what this method does.

Graphically, to do this you would mark the points x=a and x=b on the graph of f(x), then draw a straight line between them. The place where this line crosses 0 is the next x-value you would try.

Algebraically, the next number to try would be b - [(b-a)(f(b))/[(f(b)-f(a)]].

A bit horrible, but it works better than halving.

Going back to the interval (1,2) (with f(1)=-1, f(2)=2), we get this:

2 - [(2-1)(2)/[2-(-1)]] = 2 - 2/3 = 1 1/3.

f(4/3) = -0.222...<0, so the new interval is (4/3, 2).

A few more iterations of this get:

(1.4,2)

(24/17,2)

(24/17,461/290)

I've put them as fractions there for ease of keeping the accuracy, but you can put them as rounded decimals as long as you keep the whole thing in your calculator. The last interval is roughly (1.41, 1.59).

This method usually gets accuracy quicker than interval bisection.

(iii) Newton-Raphson

This one is slightly different because it only uses one value, not an interval. It's similar to linear interpolation, but instead of using another point to draw a line, it uses the gradient of one point.

That means that to use this method, the function has to be differentiable. It is also by far the quickest method of the three though, and if given the choice I definitely recommend using this one.

Graphically, the tangent to an initial point would be drawn and wherever that line crosses 0 would become the new point.

Algebraically, the next point is given by a - [f(a)/f'(a)], where f'(x) is the derivative of f(x).

Our f(x) is still x² - 2, which means that in this case f'(x) = 2x.

Taking the initial point as 1, the next one would be given by 1 - [(-1)/2] = 1.5.

Then, 1.5 - (0.25/3) = 1.41666...

Continuing this gives (as far as my calculator will go):

1.414215686...

1.414213562...

And from there it stays on that. So in just three stages, there's already an answer to 9 decimal places.

Some important things:

The functions that these are used on must be continuous (or, in the case of the first two, continuous within the initial interval). Also, the methods might start acting strangely if there is more than answer in the interval.