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In this notes sheet... Composite Structure Composite Types Critical Fibre Length Fibre Volume Fraction Loading Fibre Composites Pure materials and alloys are good for standard applications, but often a very specific set of properties is required from a material. In this instance, a composite may be used or even designed in order to maximise the best properties of a number of materials combined. A composite is a material that employs multiple different phases to attain better specific properties than either phase alone. There has to be a noticeable boundary between the two phases, and one needs to be introduced to the other rather than the two phases forming simultaneously – a two-phase slow cooled metal alloy is not a composite, then. Composites occur naturally as well as artificially. Good examples are bones and carbon fibre respectively. Composite Structure There are two main phases in a composite: the continuous matrix phase is the weaker phase that transfers applied loads to the reinforcement. It also acts as protection for the reinforcement. the dispersed reinforcement phase adds the desired property (strength, stiffness, hardness etc.) to the material by delaying crack dispersion. The matrix could be a metal, ceramic or a polymer. The reinforcement is generally a ceramic (carbon is especially common), but could take a number of forms: Particulate reinforced composites use small particles, like spheres or flakes. Fibre reinforced composites contain fibres. These could be very short, fairly long, or even continuous throughout the structure. Structural Composites may use particles, fibres, or even both for reinforcement, arranged in specific forms to maximise strength: honeycombs, layers and sheets are common. Types of Composites Metal Matrix Composites (MMCs) A cermet is a ceramic-metal composite. Typically used for cutting tools and dies. Advantages include high elastic modulus, toughness, and ductility. Disadvantages include the high density and expense. Standard matrix materials include: Aluminium & aluminium-lithium alloys Magnesium Copper Titanium Super-alloys. Standard reinforcement materials include: Graphite Alumina (aluminium oxide) Silicon carbide Boron Tungsten carbide Ceramic Matrix Composites (CMCs) Generally used for temperature and corrosion sensitive applications, like engine components and deep-sea mining, or extremely hard cutting tools. Advantages include the excellent corrosion and temperature resistance. Disadvantages are the extreme brittleness and expense. Standard matrix materials include: Silicon carbide Silicon nitride Aluminium oxide Standard reinforcement materials include: Carbon Aluminium oxide Polymer Matrix Composites (PMCs) Generally used for lightweight structures, like aircraft, vehicles, sporting, and marine equipment. Advantages include low density, easy processing, and specific properties Disadvantages include poor temperature and chemical resistance. Standard matrix materials include: Nylon Polypropylene (PP) Epoxy Phenolic Polyester Standard reinforcement materials include: Glass Carbon Boron Aramid (Kevlar) Critical Fibre Length The length of fibres makes a huge difference to the composite’s performance. Too short, and the adhesive bond between the fibre and the matrix will break before the fibre itself breaks, so the fibre pulls out of the structure Too long, and the adhesive bond between matrix and fibre transmits too much load, breaking the fibre. As with most things, a fine balance is required between the two. This is known as the critical fibre length. For a uniform circular cross-section fibre of half-length x, the shear force at the fibre-matrix boundary is given by: The force required to break the fibre is given by: The critical half-length is when these two forces equal one another: The critical length is therefore: Standard fibre lengths are: 0.2mm for a carbon fibre in an epoxy matrix 0.5mm for a glass fibre in a polyester matrix 1.8mm for a glass fibre in a polypropylene matrix Fibre Volume Fraction Fibre composites are generally described in terms of fibre volume fraction: this is (you guessed it) the proportion of the materials volume that is taken up by fibres not matrix. For fully aligned fibres, the volume fraction could be up to around 65%. Generally, it is lower. For uniform cross-sectional and continuous fibres, the area fraction is the same as the volume fraction. Loading Fibre Composites Isostrain When the fibres are parallel to the applied load, and the fibre-matrix adhesive bond does not break, the matrix and fibres experience the same strains: The total force is therefore the sum of the forces in the fibres and matrix: When both the fibres and the matrix are in the linear elastic region, Hooke’s law applies to both: Therefore: Isostress When the fibres are perpendicular to the applied stress, the stress in the fibres and matrix are the same: Therefore: Comparing Isostrain and Isostress Stress-Strain Behaviour It is harder to predict the strength of a composite without empirical investigation, however we can predict the behaviour by assuming it will initially perform similar to the fibre, but deform uniformly once the matrix begins to undergo plastic deformation.
- A-Level Maths Cheat Sheet
This is absolutely not a cheat sheet... We do not endorse cheating at all! It's just a better name than "List of vital but forgettable equations for A Level Maths" The Formula Book for Edexcel A-Level Maths and Further Maths can be seen here and downloaded here: Trigonometry Cosine Rule a² = b² + c² - 2bc cosA Area of a Triangle Area = ½ ab sinC Radians 1° = π/180 1 rad = 180/π Small Angle Approximations sinθ ≈ θ tanθ ≈ θ cosθ ≈ 1 - θ²/2 Trig Identities sin² θ + cos² θ ≡ 1 tan θ ≡ sin θ / cos θ 1 + tan² x ≡ sec² x 1 + cot² x ≡ cosec² x Angle Addition Formulae sin(A ± B) ≡ sinA cosB ± cosA sinB cos(A ± B) ≡ cosA cos B ∓ sinA sinB tan(A ± B) ≡ (tanA ± tanB) / (1 ∓ tanA tanB) Double Angle Formulae sin2A ≡ 2 sinA cosA cos2A ≡ cos²A - sin²A ≡ 2cos²A - 1 ≡ 1 - 2sin²A tan2A ≡ (2 tanA) / (1 - tan²A) R-Addition Formula a sinx ± b sinx can be expressed as R sin(x ± α) a cosx ± b sinx can be expressed as R cos(x ∓ α) Where: R cos α = a R sin α = b R = √(a² + b²) Differentiation & Integration From First Principles: Product Rule: Quotient Rule: Integration by Parts Trigonometric Integration ∫ xⁿ dx = (xⁿ⁺¹) / (n+1) + c ∫ e ˣ dx = e ˣ + c ∫ 1/x dx = ln|x| + c ∫ cos(x) dx = sin(x) + c ∫ sin(x) dx = -cos(x) + c ∫ tan(x) dx = ln|sec(x)| + c ∫ sec²(x) dx = tan(x) + c ∫ cosec(x) cot(x) dx = -cosec(x) + c ∫ cosec²(x) dx = -cot(x) + c ∫ sec(x) tan(x) dx = sec(x) + c ∫ f'(ax+b) dx = f(ax+b) / a + c Series Expansions The Binomial Expansion: The Taylor Series: The Maclaurin Expansion: Common Expansions:
- The Second Law of Thermodynamics
In this notes sheet... Reversible & Irreversible Processes The Second Law Heat Engines & Efficiency Clausius' Statement Actual vs Ideal Efficiencies Improving Efficiency The Carnot Cycle The second law is used for a number of important calculations: finding temperature differences, reversibility, and efficiency of reversible vs irreversible heat engines. Reversible & Irreversible Processes Let us take the classic example of a quasi-equilibrium expansion: If the mass on the piston is made up of a vast number of minuscule weights, like sand, removing a single grain pushes the piston up ever so slightly. The change is so small, that the return to equilibrium is almost instant. For each grain of sand removed, we have an additional point on the P-V diagram – as pressure decreases, volume increases. This process is reversible It is reversible, because if we return the grains of sand one at a time, the piston will move down by the same amount. We neglect friction or leakage. If, however, the weight consist of one large mass, and this is removed, the process is not in quasi-equilibrium. The process is irreversible In this example, the expansion itself is quasi-equilibrium, as we have returned to the grains of sand. However, when the piston hits the top or bottom stoppers, it makes a sound, releasing energy. This energy cannot be gotten back, so The process is irreversible What makes a process irreversible? Friction Unresisted expansion Heat transfer to surroundings Combustion Mixing of different fluids Clearly, then, the majority of real-life processes are irreversible. The Second Law Take the example of a spindle rotating in a container: Spindle rotates Gas in container heats up Heat from gas is conducted through container and heats up the surroundings If we reverse this: Surroundings are heated Gas inside container heats up Spindle does not start rotating We know from experience that this is the case, but why? This is where the second law comes in: Heat cannot be transferred from a cold body to a hotter body without a work input. This is the all-important idea of direction: some things are allowed backwards as well as forwards, but many things are only allowed forwards. Heat Engines & Efficiency A heat engine is a device that connects a hot and cold reservoir with a work input or output. From the first law: The second law is often also used to describe efficiency, η: Using the equation from the first law above, the efficiency can also be written as: Reversed Engine For a reversed engine, like a refrigerator, work is the input not the output. The efficiency it known as the ‘Coefficient of Performance’ (COP). In a fridge, the output is the cold reservoir, so COP is given as: Clausius’ Statement The Clausius statement of the second law tells us that it is only possible to reverse a heat engine if there is a net work input from its surroundings: A reversible engine is always more efficient than an irreversible engine. We can prove this by contradiction: if we take two engines, one reversible and one irreversible, and assume that the former has an efficiency of 10% and the latter 20%: If we now reverse the reversible engine: Modelling both engines as a single control volume: According to the Clausius statement, this is impossible. Similarly, we can use Clausius’ statement to show the 100% efficient engines are impossible: You cannot have an E100 engine Taking a 100% efficient engine and connecting its output to a reversed engine looks like this: Modelling these two engines as one control volume once again violates the Clausius statement: This is summarised by the Kelvin-Planck statement: No heat engine can deliver a work output equivalent to the heat input from a single reservoir. Actual vs Ideal Efficiencies Since no heat engine can be 100% efficient, it is often unhelpful to talk about an engine’s actual efficiency. Instead, when talking about its performance, we compare its efficiency to that of its ideal (reversible) counterpart: The efficiency of a reversible engine is a function of the temperatures between which the engine operates. This leads to the equation: This is due to the absolute temperature scale, as defined by Kelvin: the scale that sets the triple point of water (0°C) as 273.16 K. This scale is identical to the ideal gas temperature scale, from which the relationship Pv=RT is derived. This can be reworked to show that for a reversible engine, heat is a function of temperature. Therefore: Since the increment of the scale is 1 K, this simply becomes: Remember this is only the case for ideal, reversible engines. There is a huge difference in the actual efficiency of a heat engine and its efficiency compared to the maximum achievable efficiency from its reversible counterpart: The actual efficiency appears to be only 10%, but really it is 50%: Reversed Engine Back to the fridge example, the ideal Coefficient of Performance (COP) for a reversible reversed engine is given as a function of the temperatures, not heats: Improving Efficiency As you can probably tell from the numbers in the example above, efficiencies of heat engines can often be quite low. Therefore, there is a constant push to increase this to boost the performance of a system, to make it more economical, or to make it more environmentally friendly. There are a few key ways of boosting a heat engine’s efficiency: Find a lower temperature cold reservoir This is difficult, however. Generally, rivers, lakes or the ocean are used for cold reservoirs, but these are rarely lower than around 10 °C Increase the temperature of the hot reservoir Increasing the temperature difference between the two reservoirs increases the maximum possible efficiency Minimise losses Insulation, less friction etc. brings the actual engine closer to its reversible counterpart Find a use for the losses Can you use the output heat for a practical purpose, such as heating? Generally, this is difficult because the cold reservoir is around 20 °C, which is not helpful for anyone. Carnot Cycle The Carnot cycle is an unobtainable, ideal cycle of compression and expansion. An ideal petrol piston engine can be modelled as a Carnot engine. It consists of four stages: 1) Isothermal Compression There is a heat output Since the process is isothermal, there is no change in internal energy. Therefore, according to the first law, ΔQ = ΔW The piston moves down The temperature does not change, so T₁ = T₂, which equals the cold reservoir temperature 2) Adiabatic Compression There is a work input Adiabatic, so first law becomes -ΔW = ΔU The piston still moves down T₃ = the hot reservoir temperature 3) Isothermal Expansion There is a heat input Isothermal, so ΔQ = ΔW The piston moves up T₃ = T₄ = hot reservoir temperature 4) Adiabatic Expansion There is a work output Adiabatic, so first law becomes -ΔW = ΔU The piston continues to move up T₁ = T₂ = the cold reservoir temperature All four processes are fully reversible
- Entropy & The Second Law
In this notes sheet... The Clausius Inequality Entropy T-S Diagrams Combining 1st & 2nd Laws Perfect Gas Processes Isentropic/Adiabatic Efficiency Entropy & Steam As seen in the notes sheet on the second law, we can compare a heat engine’s actual efficiency with that of its reversible counterpart. If we want to compare these in greater detail, we need to use entropy. The Clausius Inequality In the notes sheet on the second law, we looked at heat engine connected to two thermal reservoirs, modelling two heat transfers only: In reality, however, this is not the case. Instead, each of these heat transfers occurs in infinitely many infinitesimally small steps, dQ: We can integrate to sum the total positive and negative heat transfers in one complete cycle: However, this on its own is not particularly helpful. Instead, we want to be able to find the heat transfer at a specific temperature. Therefore, the Clausius inequality looks at Q/T instead: Reversible Cycles The Carnot cycle is the ideal reversible cycle: Applying the Clausius inequality: Since the process is reversible, the two ratios must equal each other. Therefore, for a reversible process: Irreversible Cycles We know that the efficiency of the reversible engine is greater than that of the irreversible one, so the heat output of the irreversible engine is greater than that of the reversible: This means that But since they are both taking heat from the same hot reservoir: Therefore, plugging these into the equation derived above: We find that for an irreversible cycle: Entropy Applying Clausius’ inequality to the two reversible cycles shown above shows that the integral of dQ/T equals zero in both. This means that the change in Q/T is path independent: it is a property. This property is called entropy, S, and the change in it is given as: The units of entropy are J/K – Joules per Kelvin If we take a cycle with one reversible and one irreversible process, we know the total change in entropy must be negative (see Clausius’ inequality above): The right-hand side is the definition of change in enthalpy, so: Whenever an irreversible process occurs, the enthalpy increases. The entropy in the system itself may decrease, but then that of the surroundings will increase significantly more. Temperature-Entropy Diagrams The area under a reversible temperature-entropy curve is the heat transfer that takes place in that process: Adiabatic Process The reversible process is isentropic, as its enthalpy change equals its heat transfer, zero The irreversible process experiences an increase in entropy. Heat Addition The Process is not adiabatic, so there is a heat transfer to the system. Both the reversible and irreversible processes increase in entropy. The irreversible process increases more in entropy. Heat Rejection There is a negative heat transfer in heat rejection This means the change in entropy for a reversible process will also be negative The change in entropy for the irreversible process could be either positive or negative, depending on how irreversible it is Carnot Cycle The Carnot cycle is a rectangle on a T-S diagram: Isothermal & reversible heat rejection, so reduction in entropy. Adiabatic & reversible, so isentropic. Temperature increases. Isothermal & reversible heat addition, so increase in entropy. Adiabatic & reversible, so isentropic. Temperature decreases. Combining First & Second Laws According to the first law, for a smaller part of a reversible process: Taking the heat transfer as the area under the T-S graph and work as the area under the P-V diagram: This equation applies to both reversible and irreversible processes. Alternatively, we can rewrite the first law in terms of enthalpy: This equation also applies to both reversible and irreversible processes. Isothermal, Reversible Processes In an isothermal process, the first law becomes: Applying this to the equation for the change in entropy in a reversible process: Substituting in the ideal gas law (Pv = RT): This only applies to reversible, isothermal processes Perfect Gas Processes For an ideal gas: For a perfect gas, Cv is constant. Therefore, the integrated entropy equation becomes: This can alternatively be expressed in terms of temperature and pressure: Or in terms of pressure and volume: Isentropic Processes For an isentropic (reversible, adiabatic) process, setting the left-hand side to zero shows us that: All of these equations only apply to perfect gasses. For steam, see below. Isentropic (Adiabatic) Efficiency In an adiabatic process, such as a turbine, compressor or pump, the ideal reversible efficiency is that of the isentropic process. Turbine Efficiency For a turbine, the ideal process is a vertical line downwards on an enthalpy-entropy diagram, whereas the actual process is a diagonal: Since the end points of both processes (reversible and irreversible) are on the same isobar, there is a difference in enthalpy. This gives the isentropic efficiency: Compressor/Pump Efficiency For a compressor or pump, the directions are reversed: Therefore the isentropic efficiency is given as: Entropy & Steam T-s Diagram for Steam Similar to the vapour dome, the T-s diagram for steam is bell-shaped: The isobars inside the dome are the same as the isotherms Pressure increases diagonally from bottom right to top left To find the limits of the dome at each temperature/pressure, steam tables need to be used Interpolation is used to find the quality of the wet steam, just like dryness fraction. h-s Diagram for Steam This is an odd one. There are some similarities to the T-s diagram above, but not many: The critical point is not on the top of the curve, it is far on the left (not even shown on this diagram). This means the sub-cooled liquid region is rarely seen Isobars and isotherms are not horizontal in the wet vapour region, but are diagonal straight lines Isobars in the superheated vapour region are curved As s increases far beyond the wet vapour region, isotherms become near horizontal (the steam comes close to being an ideal gas at high entropies)
- Elasticity & Collisions
According to Hooke's Law, tension, T, is directly proportional to extension, x, in elastic strings and springs: T = kx The constant of proportionality, k, depends on two things: the modulus of elasticity, λ, of the material, and its unstretched, natural length, l: Note: natural length is noted with a lowercase 'L' - do not confuse this with an uppercase 'i'. The font of these note sheets is not massively helpful here, so sorry! Tension, T, is a force so is measured in Newtons. Extension and length are both distances, measured in metres. This gives the modulus of elasticity must also be measured in Newtons. Strings can be compressed as well as stretched. In this instance, Hooke's law still applies but the force in the spring, thrust (helpfully also written as T), acts the other way. Modelling For all questions, strings and springs are modelled as light. This means you do not take into account its weight. See the notes sheet on modelling in mechanics for single maths - the same rules apply. Elastic Potential Energy When an elastic string or spring is stretched or compressed, it stores elastic potential energy within it. This is equal to the work done in stretching/compressing the string/spring. The work done is represented by the area under a force-extension graph: Using the equation for the area of a right-angled triangle, A = ½ab, we can find the equation for the elastic potential energy stored in an elastic system: Worked Example Elastic Collisions in One Dimension Newton's law of restitution is used to solve problems where two particles collide directly: Direct impact means the the particles are moving along the same straight line when they collide. This means the collision is "head-on", so to speak. Newton's law of restitution depends on the coefficient of restitution, e, between the two colliding objects: 0 ≤ e ≤ 1 In a perfectly elastic collision, e = 1, so the two particles rebound at the same speeds In a perfectly inelastic collision, e = 0, so the particles coalesce (join together) When there are two unknowns in a problem, you can use the principle of conservation of linear momentum to solve simultaneously: Collisions with a Wall When a particle collides with a wall, the same principles apply. The only difference is, the wall does not move. Therefore, Newton's law of restitution is just taken as: Kinetic Energy The definition of a perfectly elastic collision (when e = 1) is a collision in which both momentum and kinetic energy are conserved. When e ≠ 1, the collision is not perfectly elastic, so some kinetic energy is lost. This can be calculated easily once you know the masses of each particle and their initial and final speeds: When an object collides with a wall, only the object experiences a change in kinetic energy Oblique Collisions of Spheres with Surfaces An oblique impact is when the collision does not act normally to the surfaces: While the collision may not be perpendicular to the surface, the impulse of the particle acts perpendicular to the surface. This means that the component of the velocity of the particle that acts parallel to the surface remains unchanged. v cos(β) = u cos(α) The components perpendicular to the surface are parallel to the impulse. Therefore, we can apply Newton's law of restitution: v sin(β) = eu sin(α) e is still the coefficient of restitution between the particle and the wall Dividing the second equation by the first eliminates u and v, leaving only the angles and e: tan(β) = e tan(α) Using sin² + cos² = 1 we can eliminate the angles to leave only u and v: Angle of Deflection The angle of deflection is the total angle through which the path of the particle changes. It is given by α + β Vectors Sometimes, the velocities will be given as vectors. If the surface is parallel to one of the unit vectors, this makes the question easier. To determine the angle of deflection, θ, if the velocities before and after are given as vectors u and v respectively, use the scalar product, u . v = |u| |v| cos(θ) Oblique Collisions of Spheres with Spheres In addition to collisions between spheres and surfaces, you need to be able to solve problems where two spheres collide in two dimensions: The impulse on each sphere acts along the line of centres of the two spheres Like with oblique collisions between spheres and surfaces: the component of the velocities perpendicular to the impulse is unchanged through the impact Newton's Law of restitution applies to the parallel components of the velocities the principle of conservation of momentum applies parallel to the line of centres
- Momentum, Impulse, Work, Energy & Power
As you know, momentum, p, is defined as the product of an object's mass and velocity: p = mv The Impulse-Momentum Principle When an object collides with something else, it experiences a change in momentum. This is equal to the impulse experienced by that object in the collision: I = Δmv The impulse can also be calculated as the product of the force of the collision, F, and its duration, t: I = Δmv = Ft Therefore, the units of impulse are Newton-seconds, Ns. Often, the force will have to be calculated using Newton's second law (F=ma), or SUVAT equations need to be used to calculate the duration or change in momentum. The Principle of Conservation of Momentum According to Newton's third law, for every reaction there is an equal but opposite reaction. This means that when two particles collide, both experience the same magnitude of impulse, but in opposite directions. Since impulse equals the change in momentum, the two particles experience equal and opposite changes in momentum, Δp and -Δp. Overall, these two changes in momentum cancel out: the total momentum before impact equals the total momentum after the impact Mathematically, this can be expressed as: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Work & Energy Work done is defined as the the amount of energy transferred from one form to another when a force causes movement. This is known as the work-energy principle. W = Fx Work done = force x distance moved. The force and motion must be in the same direction Because of this, you will often need to use trigonometry to find the correct components of forces: The units of energy are joules, J - so when the force is given in Newtons, N, and the distance in metres, m, the work done has units joules, J. Work Done Against Gravity & Gravitational Potential Energy Work done against gravity is given by the equation: W = mgh This can easily be derived from the work-done equation above: W = max: substitute F = ma into W = Fx W = mah: for work done against gravity, we tend to refer to the distance, x, as height, h W = mgh: acceleration due to gravity is g, 9.8 m/s² According to the Work-Energy Principle, work done is equal to energy transferred. Therefore, gravitational potential (G.P.E.) is also defined with this equation: G.P.E = mgh Kinetic Energy Kinetic Energy, KE, is given by the equation: KE = ½mv² This can also be derived from the work-done equation: a = (v²-u²) / 2s: rearrange v² = u ²+ 2as to make a the subject F = m(v²-u²) / 2s: substitute this into F = ma Fs = ½m(v²-u²): multiply both sides by displacement, s W = ½m(v²-u²): Fs = Fx = Work done ΔKE = ½m(v²-u²): Work done = the change in energy KE = ½mv² Power Power is defined as the rate at which work is done: P = Fv The units of power are Watts, W. If the force is given in Newtons and the velocity in m/s, the units of power are watts, 1 joule per second.
- Complex Numbers, Argand Diagrams & Loci
We know from single maths that if the discriminant of a quadratic equation is positive, it has two solutions, if the discriminant = 0, there is one repeated root, and if it is negative there are no roots. Wrong. If the discriminant is negative, the quadratic has two imaginary roots, given in terms of i, the square root of -1: i = √-1 A complex number is a sum of a real and an imaginary number, given in the form a + b i, where a is the real number and b the coefficient of i. Often, complex numbers are represented by the letter z. Adding and Subtracting Complex Numbers To add or subtract complex numbers, treat the real and imaginary terms separately: (a+b i) + (c+d i) = (a+c) + (b+d)i A few examples: 3 + (2+4i) = (3+2) + (4i) = 5+4i (1+2i) - 4i = (1) + (2-4)i = 1-2i (2-2i) + (6+i) = (2+6) + (-2+1)i = 8-i Multiplying Complex Numbers To multiply an imaginary number by a real constant, just expand the brackets: k(a+b i) = ka+kb i For example: 3(2+4i) = 3(2) + 3(4i) = 6+12i To multiply two imaginary numbers, expand like any quadratic, using the fact that: i² = -1 For example: (2+3i)(3+4i) = 6 + 8i +9i +12i² = 6 + 17i - 12 = -6+17i Complex Conjugation Just like conjugate pairs when rationalising a denominator, the conjugate of a complex number has the opposite sign: For any complex number z = a +bi, the conjugate is given as z* = a -bi The complex conjugate is given an asterisk. For any complex number z, the product of z and z* is a real number For example, if z = 4+5i: z* = 4-5i zz* = (4+5i)(4-5i) = 16 + 20i - 20i -25i² = 16 + 25 = 41 Solving Complex Equations Complex roots always come in conjugate pairs. Therefore, if the discriminant of a quadratic is negative, if z is one of the roots, z* will be the other root. For example, the roots of the equation x² - 6x + 13 = 0 are x = (3+2i) and x = (3-2i) Often, equations with complex roots will use z instead of x. The same rule applies to cubic and quartic graphs: If a cubic graph has just one real root, it will have two complex roots, z and z* If a quartic graph has two real roots or one real repeated root, then it will also have two complex roots, z and z* If a quartic graph has no real roots, it will have two complex conjugate pairs as roots: z, z*, w and w* All cubic graphs must have a total of three roots, real or complex All quartic graphs must have a total of four roots, real or complex Argand Diagrams Complex numbers can be represented graphically on an Argand diagram. The x-axis is the real axis, Re, and the y-axis is the imaginary axis, Im. To plot a complex number, treat it as a coordinate: The complex number z = (a + bi) has coordinates (a, b) on an Argand diagram This can be represented as a position vector, at an angle, θ, to the positive real axis (generally in radians): Like any vector, the modulus (the magnitude) can be found using Pythagoras' theorem: |z| = √(a²+b²) and tan of the angle is given as: tan(θ) = b/a Modulus-Argument Form The magnitude of the position vector for a complex number, r, is known as the modulus, and the angle it makes with the positive real axis, arg z, is called the argument. Therefore, complex numbers can also be expressed in modulus-argument form: z = r(cos(θ) + i sin(θ)) r = |z| arg z = θ Multiplying & Dividing Mod-Arg Loci A locus of points is a collection of points that can be anywhere along a particular line. These can be represented with complex numbers on an Argand diagram. The distance between two complex numbers, z₁ and z₂, is given as |z₂ - z₁| If, instead of defining both complex numbers, you just define one and give one general complex number, you get the locus of a circle: |z - z₁| = r is a circle with centre z₁ and radius r This can also be given as |z - (a+bi)| = r, where r is the radius and (a, b) the centre. You can find the locus of points for a perpendicular bisector of a line segment between two points: The perpendicular bisector of a line segment between two complex numbers, z₁ and z₂, is given as |z - z₁| = |z - z₂| Using the argument of a complex number, z₁, we can express the locus of points for a half line. This is a line from, but not including, the fixed point z₁ at an angle of arg(z-z₁): arg(z-z₁) = θ is a half line from but not including the point z₁ at an angle of θ to the horizontal. Regions Loci can be used to represent regions on an Argand diagram: Exponential Form Complex numbers can also be expressed in exponential form, z = r e^iθ, using Euler's relation: Euler's relation is derived from the Maclaurin series expansions of sin(θ), cos(θ) and eˣ. See Further Maths Notes Sheet on Series. This leads to the exponential form for a complex number, z: Where r = |z| and arg z = θ just like in modulus-argument form. Multiplying & Dividing The same rules as for modulus-argument can be applied to multiplying and dividing complex numbers in exponential form: De Moivre's Theorem Euler's relation can be used to find powers of complex numbers that are in the modulus-argument form. (r (cos(θ) + i sin(θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ)) For any integer n This can be proven with Euler's relation: It can also be proven using proof by induction (See the Notes Sheet on this) Trigonometric Identities De Moivre's theorem, combined with the binomial expansion, can be used to derive trigonometric identities. To drive an equation for sin(nθ) or cos(nθ): To drive an equation for sinⁿ(θ) or cosⁿ(θ): There are four standard results you need to know: An example of how to apply these: nth Roots of a Complex Number Just as a real number, x, has two square roots, ±√x, any complex number has n distinct nth roots. If z and w are non-zero complex numbers and n is a positive integer, the equation zⁿ = w has n distinct roots. We can use de Moivre's theorem to find the solutions to zⁿ = w by taking into account the fact that its argument is repeated every 2π. For any complex number z = r(cos(θ) + i sin(θ)), z = r(cos(θ + 2kπ) + i sin(θ + 2kπ)) Where k is any integer When zⁿ = 1 In general, the solutions to zⁿ = 1 are z = cos(2kπ / n) + i sin(2kπ / n) for k = 1, 2, 3, ..., n These are known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω such that: the nth roots of unity are 1, ω², ω³, ω⁴, ..., ωⁿ‾¹ 1, ω, ω², ω³, ω⁴, ..., ωⁿ‾¹ form the vertices of a regular n-gon 1 + ω² + ω³ + ω⁴ + ... + ωⁿ‾¹ = 0 Using nth Roots Geometrically The nth roots of any complex number, z, form the vertices of a regular n-gon with the origin at its centre The size and orientation of the polygon depend on the complex number z. If you know the coordinates of a single vertex, you can find the others by rotating that point about the origin by 2π/n. This is the same as multiplying by the nth roots of unity. If z₁ is one of the roots of the equation zⁿ = w, and 1, ω, ω², ω³, ω⁴, ..., ωⁿ‾¹ are the roots of unity, then the roots of zⁿ = w are z₁, z₁ω, z₁ω², z₁ω³, z₁ω⁴, ..., z₁ωⁿ‾¹
- Proof by Induction
Proof by induction is used to prove that a general statement is true for all positive integer values. All proofs by mathematical induction follow four basic steps: Prove that the general statement is true when n = 1 Assume the general statement is true for n = k Show that, if it is true for n = 1, the general statement is also true for n = k+1 Conclude that the general statement is true whenever n ∈ ℕ All four steps must be shown clearly in your workings: prove, assume, show, conclude. Proving Sums Often, questions will use the standard results for the sums of r, r² and r³. Regardless, the method for all sums is the same and follows the four steps above. Quick Tip For the 3rd step, it is generally best to write the last line out first using the general function - just substitute (k+1) into it. You know that this is the answer you want to reach, so use it as a target to help you. Proving Divisibility Results Again, follow the four standard steps for proof by induction. For divisibility results, make step 2 equal any general multiple of the divisor: Proof Using Matrices Exactly the same four steps apply:
Sigma notation is used to write series quickly. This describes each term of a series as an equation, where substituting in values for r gives that term in the series. There are standard sums of series formulae that are used all the time: The Sum of Natural Numbers To find the sum of a series of constant terms, use the formula: To find the sum of the first n natural numbers (positive integers): If the series does not start at r = 1, but at r = k, subtract one series from another, where the first goes from r = 1 to r = k, and the second from r = 1 to r = (k-1): Expressions in sigma notation can be rearranged to simplify complicated series: The Sum of Squares The first sum formula above, for constant terms, is linear. The second, for the sum of natural numbers, is quadratic. Therefore, the formula for the sum of square terms is cubic. To find the sum of the sum of the squares of the first n natural numbers: The Sum of Cubics Following through with this pattern, the formula to find the sum of the cubes of the first n natural numbers is quartic: The Method of Differences The method of differences is used when the expansion of a series leads to pairs of terms that cancel out, leaving only a few single terms behind. When the general term, u₁, of a series can be expressed in the form f(r) - f(r+1), the method of differences applies. This means that: To solve problems with the method of differences Write out the first few terms to see what cancels Write out the last few terms to see what cancels Add the left overs from the beginning and the end Often, you will need to use partial fractions Example The Maclaurin Series We often work with first and second order derivatives, but why stop there? Functions that can be written as an infinite sum of terms in the form axⁿ can be differentiated infinite times. We already know of a few such examples: the binomial expansion of 1 / (1-x) forms an infinite polynomial, 1 + x + x² + x³ + x⁴ + ... the binomial expansion of 1 / √(1+x) forms an infinite polynomial, 1 + ½ x - ⅟₈ x² + ⅟₁₆ x³ - ⅟₃₂ x⁴ + ... eˣ = 1 + x + ½ x² + ⅟₆ x³ + ⅟₂₄ x⁴ + ... This happens when x = 0 is substituted into increasing orders of derivatives, which, in turn, increases the power of the polynomial. This is only true if the function can be differentiated an infinite number of times, and if the series converges. When the above criteria are satisfied and each derivative has a finite value, the Maclaurin series expansion is valid: There are many standard versions of the Maclaurin series expansion that are useful to know, as they crop up a great deal:
- Roots of Polynomials
Won't lie, this topic isn't particularly fun. It's just a whole lot of algebra and equations that you need to know and get right. At least the topic is small. To be perfectly honest, the only challenge is writing everything out clearly, and not getting confused between the Greek and Latin alphabets: don't confuse a with α don't confuse b with β don't confuse g with γ don't confuse d with δ a, b, c, d & e are real constants α, β, γ, δ are roots of polynomials Quadratic Polynomials A quadratic equation in the form ax² + bx + c = 0 has two possible roots, α and β. These can be distinct and real, repeated (the same), or complex conjugates. The roots can be added or multiplied to give the following results: Squaring or cubing the roots gives different results: Cubic Polynomials A cubic equation in the form ax³ + bx² + cx + d = 0 has three possible roots, α, β and γ. The roots can be added or multiplied to give the following results: Squaring or cubing the roots gives different results: Quartic Polynomials A quartic equation in the form ax⁴ + bx³ + cx² + dx + e = 0 has four possible roots, α, β, γ and δ. The roots can be added or multiplied to give the following results: Squaring the roots gives different results: Linear Transformations of Roots If you know the sums and products of roots of a polynomial, you can find the equation of another polynomial whose roots are a linear transformation of the first. For example, if the equation x² - 2x + 3 = 0 has roots α and β, find the equation that has roots (α+2) and (β+2). This can be done in two ways: Use the roots of polynomial equations above Rearrange the roots and substitute into the first equation Method 1 (α+2) + (β+2) = α + β + 4 = -b/a + 4 = 6 (α+2)(β+2) = αβ + 2α + 2β + 4 = αβ + 2(α+β) + 4 = c/a - 2b/a + 4 = 11 So the equation is w² - 6w + 11 = 0 The coefficient of each term must be the same as the initial equation Method 2 Let w = x + 2 Rearrange for x: x = w - 2 Substitute into first equation: (w-2)² - 2(w-2) + 3 = 0 w² - 4w + 4 -2w + 4 +3 = 0 w² - 6w + 11 = 0