# FP1 (MEI) mind map

A mind map on the syllabus for Further Pure 1 for MEI. apologies that some of the bubbles are rather large and may be a little confusing without being able to express equations properly but hope it's sufficient. Enjoy!

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- Created by: Jamie MacLeod
- Created on: 20-01-13 18:08

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- FP1 (MEI)
- Matrices
- Matrix multiplication is NOT commutative. i.e. AxB does not= BxA
- For matrices to be able to be multiplied, AxB, the amount of columns in A must equal the amout of rows in B.
- The determinant of a 2x2 matrix, (a b/c d) is ab-cd. this is the area of the parallelogram made from (1 0/0 1)
- The identity matrix of a matrix is the matix that maps it onto (1 0/0 1)
- Matrices can be used to solve simultaneous equations, e.g. ax+by=c and dx+ey=f can be represented as (a b/d e)(x/y)=(c/f)
- Transformations
- a point X, (x/y) will be mapped onto a point X', (x'/y') by the matrix M (a b/c d) if X'=MX
- A rotation anticlockwise about the point O of theta degrees is (Cos(theta) -Sin(theta)/ Sin(theta) Cos(theta))
- An enlargment of scale factor 2 is (2 0/0 2)

- Complex Numbers
- If a quadratic Equation has root alpha, then the other root is alpha* where alpha* is the Conjugate pair. i.e. 1-2j, and 1+2j
- j, is the root of -1, so that j^2=-1
- When dividing a complex number, it simplifies much like rationalising the denominator, in that you multiply it by its conjugate pair.
- Loci
- the Modulus of a complex number, |a+bj|, is pythagoras of the two coefficients. ,/a^2+b^2
- the distance between z, and a complex number is |z-(a+bj)|=r where r is the distance to the point z from the complex number on an Argand diagram.
- The arugment between a point z, and complex number a+bj is arg(z-(a+bj)) where the argument is the able between the positive x horizontal.
- Modulus argument form is a way of expressing a complex number. if r is the modulus of the numer, and theta the argument, then the modulus argument is rCos(theta)+rSin(theta)j

- Quadratic equations have complex roots when the determinant of the formula<0. so b^2-4ac<0

- Curve Sketching
- The vertical asymptotes are where the denominator=0
- The horizontal asymptote is when x tends towards infinity.
- Crosses the x-axis when the numerator=0,and crosses the y-axis when x=0
- Use the graph for inequalities and substitute the according y-value to get the x-value where it crosses.

- Algebra
- Identites are used to represent two things that are always equal to each other. for the example, 2x^2-13x+25=A(x-3)^2 -B(x-2)+C, equate x^2 terms and A=2. equate x terms and -6A-B=-13 so B=1. then equate additional integers to get 9A+2B+C=25, so C=5. sometimes an x value can be substituted to cancel unknowns
- a quadratic equation, ax^2 +bx +c=0, has roots alpha and beta. alpha+beta =-b/a and alpha x beta= c/a. for a cubic equation, ax^3 +bx^2 +cx +d=0, that has roots alpha beta and gamma. alpha + beta +gamma= -b/a. (alpha)(beta) + (alpha)(gamma) + (beta)(gamma) = c/a and (alpha)(beta)(gamma)= -d/a. for a quartic equation, this continues...
- If you are wanting the sum of alpha^2 (alpha^2 +beta^2 +gamma^2) but only have the three sums that are linear, then it is (-b/a)^2-2(c/a).

- Proof by induction (using the example sum of r(3^(r-1))=0.25[3^n(2n-1)+1])
- Prove that it is true for n=1. 1(3^(1-1)=1(1)=1. 0.25[3^1(2-1)+1=0.25(4)=1
- assume it is true for n=k. 0.25[3^k(2k-1)+1].
- add the next term, k+1 and rearrange to get n=k+1. 0.25[3^k(2k-1)+1] +(k+1)(3^k). 0.25[3^k(2k-1)+1+4(k+1)3^k]. 0.25[3^k(6k+3)+1]. 0.25[3^k(3)(2k+1)+1]. 0.25[3^(k+1)(2(k+1)-1)+1].
- State your findings. It is true for when n=1, and if it is true for n=k, then it is true for n=k+1 and therefore true for all positive integers of n.

- Method of differences involves splitting the fraction into two, and putting in value to see where they cancel. put in the first few, then n-1, and n. then derive an equation from where they dont cancel.
- Finite Series
- the sum of r= 1/2(n)(n+1)
- sum of r^2= 1/6(n)(n+1)(2n+1)
- sum of r^3= 1/4(n^2)(n+1)^2

- Matrices

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