n! = n factorial = n(n-1)(n-2)(n-3)…..× 3 × 2 × 1 NOTE: 0! = 1

n! = n factorial = n(n-1)(n-2)(n-3)…..× 3 × 2 × 1 NOTE: 0! = 1
Advanced Higher AH – REVISION NOTES Unit 1 UNIT 1 OUTCOME 1 Binomial Theorem n! = n factorial = n(n-1)(n-2)(n-3)…..× 3 × 2 × 1 NOTE: 0! = 1 Binomial Co-efficient 1 Binomial Coefficient 2 Binomial Coefficient 3 ROW PASCAL’S TRIANGLE POLYNOMIAL 1 2 3 4 5 1 Row 2 Term 0 1 1 Row 3 Term 2 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Row 4 Term 3 Row 5 Term 0 Row number of Pascal’s triangle Binomial Coefficient Term number working from left to right starting at zero Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Binomial Theorem General Term
Unit 1 Binomial Theorem General Term The term INDEPENDENT of x is the coefficient where x0. (No x term) Use Binomial Theorem for numbers 1.043 = ( ) Using Pascal’s triangle = 1(1)3(0.04)0 + 3(1)2(0.04)1 + 3(1)1(0.04)2 + 1(1)0(0.04)3 = 1 + 3(0.04) + 3( ) = = Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1 Partial Fractions
TYPE 1: Distinct Linear Factors: TYPE 2: Repeated Linear Factor: TYPE 3: Irreducible Quadratic Factor is irreducible as it cannot be factorised into linear factors. First step is to show that it is irreducible using the DISCRIMINANT. therefore irreducible. Note the numerator for the irreducible factor is Bx + C. TYPE 4: Improper Rational Factor In this example notice that the degree of the numerator is NOT less than the degree of the denominator. First step must be to divide. Algebraic Long Division Now express as partial fraction. NOTE: Remember to include constant in answer Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1
DIFFERENTIATION / INTEGRATION Differentiation Integration Lanark Grammar Mathematics Department Mrs Leck

S A T C AH – REVISION NOTES Advanced Higher Unit 1 Addition Formulae
Product Formulae 0o 30o 45o 60o 90o Sin Cos Tan 1 - 90 45 S A 1 45 180 1 T C 30 270 2 60 1 Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1 UNIT 1 OUTCOME 2
A function CANNOT be differentiated at a point if it is a x y a x y NOT SMOOTH (corner) No tangent DISCONTINUOUS No tangent Rate of Change y displacement velocity acceleration Product Rule Quotient Rule Rules of Logs: Stationary Points and Points of Inflection Lanark Grammar Mathematics Department Mrs Leck

Applies to all non-zero values of x. Magnitude signs not necessary if x is positive. Only applies to linear integrals eg. not Integration by substitution NOTE: When we use substitution to create a definite integral the limit MUST be expressed in terms of the new variable. If the integrand contains a term in the form , it is often easier to use a trig term. Volume of Revolution about x axis Volume of Revolution about y axis NOTE: Find y2 or x2 and substitute before integrating. π can be taken outside of the integration as it is constant. When INTEGRATING with respect to time: Acceleration Velocity Displacement Acceleration a(t) Velocity v(t) = Displacement s(t) = Lanark Grammar Mathematics Department Mrs Leck

all elements of B are used
Advanced Higher AH – REVISION NOTES Unit 1 UNIT 1 OUTCOME 4 ONE TO ONE Function Each value of f(x) in the range is produced by ONLY one value of x from the domain. MANY TO ONE Function Function maps more than one element in the domain to the same element in the range. ONTO FUNCTIONS Range = Codomain r 4 r 4 8 h 2 h 2 s 1 5 q q 5 A B ONTO all elements of B are used A NOT ONTO 8 and 1 in Set B are not used B Inverse Functions The graph of a function is found by reflecting it in line y = x. Find the inverse by interchanging x and y in the formula and then make y the subject of the formula. eg. Interchange x and y Make y the subject of the formula Modulus Function The modulus of a function takes the POSITIVE values of x. The function does NOT appear below the x-axis. EVEN – symmetry about the y-axis. ODD – symmetry about the x-axis. Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1 Sketching Asymptotes
STEP 1 – ODD / EVEN Check as it simplifies drawing it if there is symmetry about an axis. STEP 2 – VERTICAL ASYMPTOTE (Check behaviour) Denominator = zero STEP 3 – NON-VERTICAL ASYMPTOTE Set x to infinity to find y = When the numerator has a degree ≥ degree of the denominator – use long division. Then set x to infinity. Gives a sloped asymptote of the form y = mx + c. STEP 4 – CROSSES X & Y AXES Find when x = 0 and y = 0. STEP 5 – STATIONARY POINTS Find and STEP 6 – NATURE OF STATIONARY POINTS Use either a nature table or find STEP 7 – SKETCH GRAPH IDENTIFYING POINTS AND ASYMPTOTES Lanark Grammar Mathematics Department Mrs Leck

1. A unique solution eg. 2. A non – unique solution eg or each of which satisfies all 3 equations. 3. No solution… there are no values which satisfy all 3 equation. There are three possible solutions: 1. Unique Row 1 or r1 Row 2 or r2 Row 3 or r3 Aim: to form 2. Non-unique If we substitute this into row 1, we get Hence various solutions are possible provided eg. The system of equations is inconsistent Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1 3. No solutions
Row 3 yields No solutions . The system of equations is inconsistent Ill Conditioning In practice, systems of equations may come about as a result of an experiment where measurements have been taken. Measurements are only as accurate as the instrument being used allows. When Gaussian elimination is used to solve these equations we get a solution were measurements made to the nearest tenth of a millimetre. A second person obtained three slightly different results of However when these three values are used as the vector on the right hand side of the system of equations, we get the solution A significant change! This is an example of ill-conditioning. “Small uncertainties in the data make large uncertainties in the solution” Lanark Grammar Mathematics Department Mrs Leck

Differentiation of an Inverse function Example: Given f(x) = x3 find f’(x) and state the derivative of f-1( x). Quick method: Differentiation of an Inverse Trig Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1 Implicit Differentiation
Differentiate x and y at the same time with respect to x. Therefore the derivative of y = The equation defines y implicitly in terms of x. Find an expression for in terms of x and y. Implicit Differentiation – Equation of the Tangent In order to find the equation of a tangent at a point differentiate the equation substitute the point into the differentiated equation to find the gradient substitute the gradient and point into the equation Implicit Differentiation – Second Derivative Find the first derivative and then differentiate again. Logarithmic Differentiation Lanark Grammar Mathematics Department Mrs Leck

AH – REVISION NOTES Advanced Higher Unit 1 Parametric Differentiation
A parametric equation is where the x and y coordinates are both written in terms of another letter. To find the gradient of the curve: Steps for Parametric Equations 1) Take the two equations for x = and y = and find the derivative of each in terms of t. 2) To find the gradient use divided by the inverse of Equations of Tangents can also be deduced using Parametric Equations. Calculate the derivative in order to find the gradient. Using the defined parameter, substitute into the parametric equations to find the coordinate (x, y) Substitute into the equation of a line. The second derivative of Parametric Equations Find the first derivative – differentiate again and multiply by the inverse of the x derivative. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 2 OUTCOME 2 Inverse Trig Functions
NOTE: It only applies to SIN and TAN. NOT COS !!!!!! Integration by Parts To integrate two functions which are multiplied together use integration by parts (diff – product rule) 1st Order DE – Separable Variables Separate out the variables. Integrate both sides with respect to the appropriate variable. 1) Find the general solution of the differential equation General Solution contains constant C. Particular Solution – Constant term C is found. NOTE: if an algebraic fraction – if the numerator has the same or greater power than the denominator – USE LONG DIVISION and partial fractions to integrate. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 2 OUTCOME3 Complex Numbers Complex Number Real
Part Imaginary Part Add, Subtract and Multiply Complex Numbers Complex Conjugates Division of Complex Numbers Multiply both numerator and denominator by the complex conjugate of the Denominator. Argand Diagrams (Geometric Representation) Modulus z r b Argument θ θ a Polar Form θ must lie between -180o and 180o. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher Multiplication / Division of Polar Form
De Moivre’s Theorem Roots of Complex Numbers 2 roots – roots lie 180 degrees apart. (360 ÷ 2) 3 roots – roots lie 120 degrees apart (360 ÷ 3) n roots – roots lie (360 ÷ n) apart. Roots of Polynomial Equations Use Algebraic Long division to find root. If the root is non real eg. a + ib then its conjugate a – ib is also a root. Trigonometric Identities Use Binomial Theorem to expand and find the real part and the imaginary part. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher Locus in the Complex Plane
The locus of z = path of the complex number z. 4 - 4 Inside the circle Circle Centre (a, b) Radius r. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 2 OUTCOME 4
Sequence – an ordered list with a fixed rule. Series – a sum where the terms form a sequence. Sequence and Series SUMMARY Standard Formulae a = constant 1) 2) 3) Sequence - an ordered list of terms Arithmetic Sequence eg. 1, 4, 7, 10,… Geometric Sequence eg. 3, 12, 48, 192,… a = first term d = common difference a = first term r = common ratio Series - sum of a sequence of terms Arithmetic Sequence eg … Geometric Sequence eg … Infinite Geometric Series Sum to Infinity = ONLY when -1 < r < 1. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 2 OUTCOME 5 Proofs
If the number x is even it can be expressed as If the number x is odd it can be expressed as If then a is divisible by b. A rational number can be represented as where p and q are integers with no common factors. COUNTER-EXAMPLE A counter-example is an example that demonstrates that a conjecture is not true. PROOF BY EXHAUSTION Examine every possible case that applies to the conjecture, proving the conjecture explicitly. PROOF BY CONTRADICTION Assume that the negation (opposite) of the statement is true By a series of steps arrive at a contradication Since all steps are valid the assumption must be false If the negation is false, the original statement must be true. PROOF BY INDUCTION 1. Check that the statement is true for some positive integer a eg. 1. 2. Assume that the statement is true for n = k. 3. Prove that the statement is also true for n = k + 1. 4. Conclude that since it is true for n = a then it is true for n = a+ 1 and it is true for n = a + 2 and so on. Thus by induction it is true for all Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 3 OUTCOME 1 Vectors
Review Higher work and definitions. Scalar or Dot Product - For two vectors a and b the scalar product is defined as a.b where θ is the angle between a and b, 0 ≤ θ ≤ 180o. If a and b are perpendicular then a.b = 0. Component of a Scalar Product - Angle between Vectors – use the scalar product. Perpendicular Vectors – when a.b = 0. Properties of Vectors – for vectors a and b, a.b = b.a. for vectors a, b and c, a(b + c) = a.b + b.c Symmetric Form Line L u line L is expressed in SYMMETRIC FORM Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher Parametric Form
The Equation of a line in symmetric form is Intersection of two line To find the point of intersection of two line in 3D space, both lines must be expressed in PARAMETRIC FORM and equate the x, y and z values. Angle between Two Lines STEP 1: Find a vector a in the direction of the line L1. STEP 2: Find a vector b in the direction of the line L2. STEP 3: The angle, θ, between the line L1 and L2 is the angle between the direction vectors a and b. EXPRESS DIRECTION VECTOR IN SYMMETRIC FORM. Equation of a Plane Vector Product Step 1 – Write the vector in component form Step 2 – Cover up row 1 and cross multiply rows 2 and 3. This gives coefficient i. Step 3 – Cover up row 2 and cross multiply rows 1 and 3. This gives coefficient -j. Step 4 – Cover up row 3 and cross multiply rows 1 and 2. This gives coefficient k. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher Vector a × b is always PERPENDICULAR to each of vectors a and b. Note: a × b is NOT EQUAL to b × a. The Angle Between Two Planes n1 is the vector normal to the plane π1 n2 is the vector normal to the plane π2 The Angles Between a Line and Two Planes If L intersects the plane π . a is the vector in the direction of line L n is the vector normal to the plane π. θ is the angle between the line L and the plane π. The Distance from a Point and a Line Line L Q P The shortest distance from point P to the line L is the PERPENDICULAR distance from P to the line. Vector is perpendicular to a vector in direction Line L. The Distance from a Point to a Plane P The shortest distance from point P to a plane π is the PERPENDICULAR distance from P to the line. Plane π Q Vector is parallel to a vector normal to the plane π. The Line of Intersection of Two Planes Two planes which intersect, do so in a LINE. To find the equation of the line of two planes intersecting – PARAMETRIC FORM. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher (m × n) × (n × p) = (m × p) UNIT 3 OUTCOME 2 Matrices
SCALAR MULTIPLICATION OF MATRICES If k is a scalar (number), the matrix kA is formed by multiplying each entry of the matrix A by k. TRANSPOSE OF A MATRIX AT The transpose of a matrix A is denoted by A’ or AT. The rows and columns have been swapped. 1) Note: A = 2 × 3 and AT = 3 × 2 SYMMETRIC AND SKEW-SYMMETRIC MATRICES - Only applies to SQUARE MATRICES. Matrix A is symmetric if AT = A. Matrix A is skew-symmetric if AT = -A. Note: the matrix is symmetrical about the leading diagonal from top left to bottom right. Note: the matrix is symmetrical about the leading diagonal which is equal to zero. MULTIPLICATION OF MATRICES 1) 2) (m × n) × (n × p) = (m × p) Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher det A = ad – bc A.A-1 = I and A-1.A = I.
IDENTITY MATRIX (I) DETERMINANTS OF SQUARE MATRICES 2 × 2 Matrix det A = ad – bc 3 × 3 MATRICES Example 1 1) Multiply the first entry of top row by the determinant of the 2 × 2 matrix remaining when the row and column are deleted. 2) Multiply the second entry of top row by the 3) Multiply the third entry of top row by the 4) Add the answers together in a pattern. INVERSE OF A MATRIX A.A-1 = I and A-1.A = I. If AB = I then B = A-1. INVERSE OF A 2 × 2 MATRIX Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher INVERSE OF A 3 × 3 MATRIX
Elementary Row Operations (ERO) As with 2 × 2 matrices, there will only be an inverse matrix if the determinant is a non-zero. Check the steps out following Gaussian steps. TRANSFORMATION MATRICES is the matrix associated with reflection in the x – axis. is the matrix associated with reflection in the y – axis. is the matrix associated with reflection in the origin. is the matrix associated with reflection in the line y = x. is the matrix associated with ANTICLOCKWISE rotation about the origin. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 3 OUTCOME 3 Further Sequence and Series
MACLAURIN SERIES APPROXIMATE ROOTS For a linear recurrence relation to converge, tending to a limit L, it must be that -1 < a < 1 Staircase and Cobweb Diagrams Diagrams may be used to find the convergence limit or Fixed Point of the recurrence relation. There are two types of diagram 1. Staircase Diagram – when a > Cobweb Diagram – when a < 0 Conditions for Convergence Whether an iterative process with converge or not depends on the starting value used and the particular formula. It is possible to determine whether an iterative process will converge or not. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 3 OUTCOME 4 Further Linear Differential Equations
A first order differential equation can be written in the form P(x) and f(x) are functions of x only. (does not include y2, ey, siny etc.) To find the general solution, first find the INTEGRATING FACTOR I(x). INTEGRATING FACTOR Once the integrating factor has been found it is then used to calculate the general solution Steps to find the General Solution: 1. Write in the form and identify P(x) = f(x) = 2. Find 3. Substitute I(x) into 4. Integrate 5. Rearrange to solve for y. NOTE: the form must be singular and not Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher Second Order Differential Equations
A second order D.E. is expressed a, b and c are constants where a ≠ 0. f(x) is a function of only x. IF f(x) = then it is HOMOGENEOUS IF f(x) ≠ then it is NON-HOMOGENEOUS. Homogeneous 2nd Order Differential Equations To find the General Solution: 1. Form a quadratic “auxiliary equation”. 2. Find the roots of the auxiliary equation. 3. The nature of the general solution depends on the roots: A) If the roots are real and distinct (b2 – 4ac > 0) where k = p and k = q and p ≠ q then A and B are constants. B) If the roots are real and equal (b2 – 4ac = 0) where k = p and p = q then A and B are constants. C) If the roots are complex conjugates (b2 – 4ac < 0) where k = p ± qi then A and B are constants. Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher Non-Homogeneous 2nd Order Differential Equations
where f(x) is a simple function. Steps to solve Non Homogeneous DE 1. Find the general solution of the homogeneous d.e. This is called the COMPLEMENTARY FUNCTION. 2. Find a particular solution of the non-homogeneous. This is called the PARTICULAR INTEGRAL. 3. GENERAL SOLUTION OF NON-HOMOGENEOUS = COMPLEMENTARY FUNCTION + PARTICULAR INTEGRAL Finding the Particular Integral Lanark Grammar Mathematics Department Mrs Leck

Advanced Higher UNIT 3 OUTCOME 4 Further Proofs Check number bases.
EUCLIDEAN ALGORITHM Example 1 Find the greatest common divisor of 140 and 252. Solution Step 1 – Express the larger number (252) in terms of the smaller number (140). 252 = 1 × Step 2 – Express 140 in terms of = 1 × Step 3 – Express 112 in terms of = 4 × The remainder is now zero and the greatest common divisor is the last non-zero remainder. Hence (140, 252) = 28. Example 2 Use the Euclidean algorithm to show that the (149, 139) = 1. Hence find integers x and y such that 149x + 139y = 1 Solution Lanark Grammar Mathematics Department Mrs Leck

n! = n factorial = n(n-1)(n-2)(n-3)…..× 3 × 2 × 1 NOTE: 0! = 1

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n! = n factorial = n(n-1)(n-2)(n-3)…..× 3 × 2 × 1 NOTE: 0! = 1

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Presentation on theme: "n! = n factorial = n(n-1)(n-2)(n-3)…..× 3 × 2 × 1 NOTE: 0! = 1"— Presentation transcript:

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