#1

Determinants.

2x2 Matrix A

a b
c d

3x3 Matrix B

a b c
d e f
g h i

|B| = a(ei - fh) - b(di - fg) + c(dh - eg)

|B| = 0 => matrix is inversible or points are coplanar or directions are linearly dependant.

Dot/Cross/Triple-Scalar Products

Dot Product
a.b = [email protected]

(a,b,c).(x,y,z) = ax + by + cz
a.b = 0 => directions are perpendicular

Cross Product
axb = [email protected]
= area of parallelogram with sides a, b

(a,b,c)x(x,y,z) =
| i j k |
| a b c |
| x y z |
= i(bz - cy) - j(az - cx) + k(ay - bx)

Triple Scalar Product
a.(bxc) =
| a1 a2 a3 |
| b1 b2 b3 |
| c1 c2 c3 |

triple scalar product gives the area of a parallelepiped, (if it equals 0 it shows the points are co-planar)

Linear Combinations

(%,\$,£)(A) = (p)

%(a) + \$(b) + £(c) = p1
%(d) + \$(e) + £(f) = p2
%(g) + \$(h) + £(i) = p3

solve for the ratios %:£ to find the combinations of the vectors to get the point.

Matrix Multiplication

A=
a b c
d e f
g h i

B =
j k l
m n o
p q r

AB =
(aj + bm + cp) (ak + bn + cq) (al + bo + cr)
(dj + em + fp) (dk + en + fq) (dl + eo + fr)
(gj + hm + ip) (gk + hn + iq) (dl + ho + ir)

BA =

(ja + kd + lg) (jb + ke + lh) (jc + kf + li)
(ma + nd + og) (mb + ne + oh) (mc + nf + oi)
(pa + qd + rg) (pb + qe + rh) (pc + qf + ri)
0
13 years ago
#2
Proof by induction:

The general method:
A proof by induction consists of showing two things:
(1) That if a theorem is true for a typical integral value of n, such as n=k, then it is also true for n=k+1.
(2) The theorem is true for a trivial value of n such as n=1 or n=2.
It is then possible to state that the theorem is true for n=1+1=2, 2+1=3… and so on for all positive integral n.
Note that in some cases, such as proving properties about odd or even numbers we instead show that if true for n=k then it is also true for n=k+2 and that it is true for a trivial odd or even number, as applicable.

Notes:
(1) If a proof involves proving a theorem regarding the sum from 1 to 2n, then when finding the expression for the summation to 2(k+1) you need to add the terms for both r=2k+1 and r=2k+2.
(2) When proving divisibility care should be taken as to whether it has been stated that n is odd or even.
(3) In these cases when trying to prove g(n)>f(n) you may find g(k+1)>f(k+1)+[Another positive function]. You can simply state that as the extra function is positive g(k+1) is certainly greater than f(k+1) and hence prove if true for n=k then it is also true for n=k+1.
(4) It should be noted that if a recurrence formulae involves more terms in n than U(n+1) and U(n) then it may be necessary to assume the given statement is true for n=k and n=k+1 or even n=k+2 in order to use the recurrence formulae given.
When evaluating trivial values of n you then need to show it is true for n=1 and n=2 or even n=3 and then you can safely generalize.
0
13 years ago
#3
Complex Numbers

Things you should know:
eit = cost + i sint
z = r(cost + i sint) = r.eit, where t=argz and r=|z|.
zn = rn(cost + i sint)n = rn(cos(nt) + i sin(nt)). This is De Moivre's theorem.
zn + z-n = 2 cos(nt)
zn - z-n = 2i sin(nt)

cos(iz) = cosh(z)
sin(iz) = i sinh(z)

|z/w| = |z|/|w|
|zw| = |z| |w|
arg(z/w) = argz - argw
arg(zw) = argz + argw

Roots of Unity
1 = cost + i sint => t=0, 2pi, 4pi, ...
1^(1/n) = cos(t/n) + i sin(t/n), where t=0, 2pi, 4pi, ...

So, the nth roots of unity can be written as:
1, w, w2, w3, ..., wn-1, where w = cos(2pi/n) + i sin(2pi/n).
Example:
The fourth roots of unity (n=4; 1^(1/4)) are given by:
1
cos(2pi/4) + i sin(2pi/4) = i
cos(4pi/4) + i sin(4pi/4) = -1
cos(6pi/4) + i sin(6pi/4) = -i

Roots of Complex Numbers
The nth roots of any complex number z are given by:
z1/n = r1/n {cos[(t + 2kpi)/n] + i sin[(t + 2kpi)/n]}
Unless a range of values is specified for t, k=0, 1, 2, ..., n-1. If t was allowed to be negative, then you can let k start from negative values, e.g.
You want to find the 3rd roots of z, and you found that z^(1/3) = 4.e^i[(pi/3) + 0.5kpi]. You're given that -pi < t <= pi, so:
k = -2, -1, 0, 1, 2. Because if k=-3, then t = (pi/3) + 0.5kpi = -7pi/6 < -pi, which is out of the range of allowed values, but k=-2 gives t = -2pi/3 > -pi, which is allowed.

Loci
Basic loci:
|z - (a+bi)| = k is a circle center (a, b) with radius k.
arg[z - (a+bi)] = t is a half-line starting from (a, b) and making an angle t with the horizontal there.
|z - (a+bi)| = |z - (c+di)| is the perpendicular bisector of the line joining (a, b) and (c, d).
arg[{z - (a+bi)}/{z - (c+di)] = t is an arc of the circle passing through points (a, b) and (c, d) such that the angle suspended on the circumference of the circle by the line connecting these points is equal to t. For negative t, simply draw the locus as if it was for positive t, then the locus you want is the reflection about the line connecting (a, b) and (c, d).
If you're stuck you can simlpy substitute z=x+iy and find an equation of the locus using algebra, but be warned that sometimes a lot of work is required.

Transformations
The genearl method is to use the information you're given in the question, e.g.
If you're given a relation: w = f(z), and then you're told that |z|=2, then you should write z in terms of w then take the modulus of both sides of the equation.

Another common way to approach a question is to write w and z in complex number form and equate real and imaginary parts, then use the information given. Remember that you can write a complex number in basic a+bi form, exponential form or modulus-argument form. E.g.
w = f(z) is a transformation that maps z onto w. Show that if:
(a) w lies on the real axis, the point representing z maps a circle;
(b) w lies on the imaginary axis, the point representing z maps a straight line.
The way I'd approach the question is by writing w=u+iv and z=x+iy, and then I'd simplify the equation into a form similar to:
w = u + iv = g(x, y) + i h(x, y)
Then I'd equate real and imaginary parts, i.e. u=g(x, y) and v=h(x, y).
(a) Let's use the information given in the question. If w lies on the real axis, then Im(w)=0, i.e. v=h(x, y)=0. Simplifying this should give us an equation for y in terms of x, and if our working was correct, the equation would be that of a straight line.
(b) Similarly, Re(w)=0, so u=g(x, y)=0, etc.
0
13 years ago
#4
Step by step approximation techniques

There are many types of differential equations which cannot be solved analytically. In order to approximate to the value of y at a certain value of x we can apply step by step methods.

The main formulae we use are:
(1) (dy/dx)0~[y1-y0]/h
(2) (dy/dx)0~[y1-y(-1)]/2h
(3) (d^2y/dx^2)0~[y1-2y0+y(-1)]/h^2
Here 'h' indicates the step length and y-1, y0, y1 are successive values of y. If y0=f(a), y(-1)=f(a-h) and y1=f(a+h)

To solve a typical equation involving only (dy/dx), x and y we normally use approximation (1). We can obtain a more accurate approximation by using approximation (2) but more information is needed and so the first approximation may need to be used so we can find an approximation for y1 to use in the second.

Whenever we have a differential equation involving (d^2y/dx^2),(dy/dx),y and x we tend to use approximations (3) and (1or2), substituting them in place of (dy/dx) and (d^2y/dx^2) respectively to obtain the iteration formulae that we desire. In this case we may be given the values of y and (dy/dx) at a certain value of x. Here we need to equate values with the approximations to obtain simultaneous equations and hence an approximation for values of y1 and y-1.

Note that step by step methods merely involve setting up an iteration formulae and using it repeatedly to approximate to values of y at particular values of x. It may be helpful to substitute in the required approximations then make y1 the subject of the formulae to ease later calculation.
Note that to give the approximation a more conventional appearance we may write y0 as yn, (dy/dx)0 as (dy/dx)n, x0 as xn, y1 as y(n+1) and so on... This makes it easier to see how the approximation may be repeated for different values of x and y.
0
13 years ago
#5
thanx guys!!!! u all helped a lot!! ^.^
0
13 years ago
#6
thank you for the help! ill post if i ever come round to writing up my own notes. reps coming ur way... eventually
0
13 years ago
#7
I decided to make some P6 notes while at the same time learning to type maths better. They might not be any use but I may as well attach them.
0
13 years ago
#8
Thanks a lot for all of these.

I spent too much time revising for P5 and kind of let everything else slip which was a little stupid. I need all the help I can get for this module.
0
13 years ago
#9
(Original post by master-chafe)
Thanks a lot for all of these.

I spent too much time revising for P5 and kind of let everything else slip which was a little stupid. I need all the help I can get for this module.
same but for chemisrty am startiung to worry, hmm should be revising
0
13 years ago
#10
(Original post by Gaz031)
I decided to make some P6 notes while at the same time learning to type maths better. They might not be any use but I may as well attach them.
I liked your notes very much - thanks for them !!

The layout also looks very good - what programme did you use to write them up?

Thanks again,

Daniel
0
13 years ago
#11
(Original post by danguetta)
I liked your notes very much - thanks for them !!

The layout also looks very good - what programme did you use to write them up?

Thanks again,

Daniel
I used latex to write them. I'm still new to using latex so writing some notes gave me a reason to learn some more of the features such as matrices.
I got started from here but there are many tutorials. It's seems difficult at first but really isn't that hard to use. http://www.artofproblemsolving.com/L...PS_L_About.php
0
12 years ago
#12
Bumped for this year's examinees
0
12 years ago
#13
Thanks a lot
0
12 years ago
#14
Hey, has anyone finished P6 yet? It's a beast of a book, we're on the vectors section.
0
12 years ago
#15
We're still on Matrices We'll only be finishing it in mid May I agree, its one huge book!
0
12 years ago
#16
Meh, I do OCR, seems your FP3 is very similar to our FP1, because we do all the matrix, proof by induction, and the basic bits of complex numbers in that module. But that's beacause OCR are stupid and wanted people to be able to do FP1 without C3 or C4
0
12 years ago
#17
Yeah I've heard about that. But don't you think its better that way, cos ppl can do AS Further Maths (FP1 + 2 applied) in their AS year as well?

We had to do A2 maths in Year 12 and then do all 6 units of FM in Year 13...
0
12 years ago
#18
(Original post by ArVi)
Yeah I've heard about that. But don't you think its better that way, cos ppl can do AS Further Maths (FP1 + 2 applied) in their AS year as well?

We had to do A2 maths in Year 12 and then do all 6 units of FM in Year 13...
I suppose it would be if we did it that way, but we're doing it the same way as you. My main problem with it is that there is no calculus in it, which is my favourite area of pure maths, and that meant that I was bored solid with matrices. Basically they just bung all of the boring stuff into one module...
0
12 years ago
#19
(Original post by rpotter)
I suppose it would be if we did it that way, but we're doing it the same way as you. My main problem with it is that there is no calculus in it, which is my favourite area of pure maths, and that meant that I was bored solid with matrices. Basically they just bung all of the boring stuff into one module...
That can't be too fun... I love calculus too
0
12 years ago
#20
If you were to do the A Level like that to get AS Maths and F.Maths you'd need 3 applied modules... if your did all Machanics you'd struggle so bad on M3.
0
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