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Maths Y13 Pure
Maths Autumn Y13
| Question | Answer |
|---|---|
| What is the range of a function? | All the values that can come out of a function (acceptable outputs) |
| What is the domain of a function? | All the values that go in (acceptable inputs) [domaIN] |
| What does the fancy Z with superscript + mean? | Positive integers not including zero (aka counting numbers) |
| What does the fancy R mean | Real numbers |
| What does the fancy Q mean | Rational numbers |
| How to add conditions to set notation | Commas (however its said by Ms Wade that you cant do epsilon R when > or < involved as it cant be any real number - feels off though) |
| What is one-to-one mapping | Every value on the domain is mapped onto a unique value in the range (e.g. general rule 2x + 5) - no crossover of graph (e.g. just straight line) |
| What is one-to-many mapping | A single value in the domain is mapped onto 2+ *unique* values in the range (e.g. two lines on graph) |
| What is many-to-one mapping | 2+ values in the domain are mapped onto the same value in the range (e.g. x^3 graph) |
| What is many-to-many mapping | Each value in the domain mapped onto non-unique multiple values (e.g. circle graph) |
| Useful tool for seeing stretches | Writing out how (x,y) would be affected |
| Drawing weird stretches on sin/cos/tan graphs | Dont stress about the scaling - the labelling is the important part |
| Multiple transformations upon the same dimension | The y transformations affect the whole equation as normal The x transformations affect only the x - not everything in the brackets. |
| Remember with scaling x by 3 | That means it becomes 1/3 x |
| 'One way stretch' | Just means stretch |
| Do the compound angle formulae have to be in radians | No |
| How to prove tan(A+B) | Divide by cosA cosB Look where you're going with the 1 to work it out |
| When working with simplifying sin, cos, etc with real values | Can turn e.g. sin(pi/6) into nice numbers first |
| If Q asks for a tanx value | Dont give the x value |
| cotangent (cot), cosecant (cosec), secant (sec) way to remember | 3rd letter is the function its the inverse of Being the inverse of a function means that its the opposite flip for sohcahtoa, and can also just be the value of tanx then ^-1 |
| Alternate method to triangle for cosx = 3/5 being converted to sinx | Sub in to sin^2 + cos^2 = 1 |
| When finding secondary solutions to a function | CHECK WITH GRAPH ALWAYS FOR WORKING MARKS YOU NEED TO DRAW IT ALSO NEED TO CHECK IF IT EVEN HAS SECONDARY SOLUTIONS - DOES IT HAVE THE SAME Y VALUE FOR MULTIPLE X POINTS IN A PERIOD |
| Generally when simplifying both sides by dividing by smth when doing sin/cos/tan | Reconsider dividing by sin as it'll just lead to messy cots rather than tans - still works but kinda annoying |
| Dividing by a variable | NEVER divide by a variable (unless forming an identity?) - loses solutions. Factorise out instead |
| Sin curve translation | A sin curve translated parallel to the x axis and stretched parallel to the y axis will have the form rsin(x + alpha) for some alpha and r. Using compound angles, can be written as asinx + bcosx |
| e.g. Put 3sinx + 4coxx in the form Rsin(x+a) giving a in degrees to 1dp Finding R | Expand Rsin(x+a) = Rsinxcosa + Rcosxsina Compare coefficients: 3 = Rcosa 4 = Rsina Square both coefficient comparisons, then add. Factorise out into R^2(cos^2 + sin^2) = R^2 = 9 + 16 R = 5 |
| e.g. Put 3sinx + 4coxx in the form Rsin(x+a) giving a in degrees to 1dp Finding a and giving equation | Using the fact Rsina/Rcosa = tana 4/3 = tana a = 53.1 degrees (principal easiest if they dont mind which solution) Sub into equation |
| Anything with e.g. theta - 68 degrees as the solution to a cos^-1 | CHANGE THE RANGE Work out the solutions before adjusting back to pure theta, changing all the solutions |
| When you're doing the R method of changing double angle formulas and they dont give you an end point | Just choose one - they all work as versions of each other |
| cos^-1(x) | Doesnt mean cosx to the power of -1 because history, instead meaning arccos Reciprocal doesnt mean inverse for trig |
| For when you have a split range where no number is in both | Use 'or' or union. Comma and 'and' not valid |
| For easy tan drawing | Draw asymptotes first |
| Multiple equals sign in one line | Can apparently only do one per line or incorrect |
| Finding the new identities | Divide the sin^2 + cos^2 identity by sin^2 or cos^2 Always derive to make sure its correct |
| e.g. 1/tantheta = 0 | Cant divide by zero so no reciprocal solution. Either show graph and select points when cot(theta) = 0 [which is where the asymptotes of tan are] or do cos/sin = 0, so cos = 0 |
| Double angle proof step 1 | Suppose we had a line length 1 projected at angle A+B above horizontal. Angle A forms right angled triangle, then the angle B forms a right angle triangle with the hypotenuse equalling 1. Line connecting from base to very top point is line XY = sin(A+B) |
| Double angle proof step 2 | Add a new right angled triangle to the gap between the previous two. XY = the two vertical lines to the right of it added together |
| Double angle proof step 3 | Find lengths of top triangle with hypotenuse 1. O = sinB. A = cosB |
| Double angle proof step 4 | Find lengths of two smaller vertical lines, each in terms of both sin and cos. Add together and equal it to sin(A+B) |
| Using double angle proof for sin(A-B) | sin(A-B) = sin(A+(-B)) = sinAcos(-B) + cosAsin(-B) Draw graphs to show cos(-B) = cosB and sin(-B) = -sinB Sub in |
| Using double angle proof for cos(A+B) | cos(A+B) just shifted by pi/2 of sin(A+B) cos(A+B) = sin(pi/2 - (A+B)) = sin((pi/2 - A) + (-B)) = sin(pi/2 - A) cos(-B) + cos(pi/2 - A) sin (-B) = cosA cosB - sinA sinB |
| Inverse trig identities | Not the same as ^-1. We have to restrict domain so one-to-one (only type of function we can inverse). Reflected in y = x. Careful that they dont end at the intersection: continue on to pi/2 after meeting at 1,1 |
| Given y = arccosx, express arcsinx in terms of y | y = cos^-1(x) x = cosy x = sin(pi/2 - y) [this is just true: reflection and shift of 90 degrees ] pi/2 - y = arcsinx |
| sin(5(x-30)) if translation happens first | Translate by 150/0 then stretch horizontally by 1/5 |
| Map graph of y = 2(x+1)^2 + 5 onto y = x^2 | ONTO x^2 so opposite direction |
| How to do inverse function | f(x) = 2x + 1 Let 2x + 1 = y [this is because reflected in y=x] y = 2x + 1 (y-1)/2 = x. then turn the y into an x and the x into an f(x) [domain and range being swapped] |
| Domain and range of inverse function | Swapped |
| Self-inverse functions | If and only if ff(x) = x |
| What does f:x ---> mean | f(x) = |
| What functions can have an inverse | A function can have an inverse if and only if it is a 1-to-1 function |
| Modulus functions | Sketches rlly useful Can do with just common sense of flipping gradient, but better to get it into the form I2x+1I < 7 (modulus only thing on that side), then turn that into -7 < 2x+1 < 7. Check graph to see if split < > |
| 'Solve the equation' | x values |
| If(x)I | Flipped in x axis |
| f(IxI) | As if the x value inputs were the positive: reflect the positive x value graph in the y axis |
| Section of curve with d^2y/dx^2 > 0 | Concave upwards (i.e. if points at 2 ends connected by line, curve below that line Vice versa for <0 |
| Points of inflection | When d^2y/dx^2 = 0, where it switches from concave upwards to concave downwards. Points of inflection can be stationary but not always. Not all d^2y/dx^2 = 0 are points of inflection |
| Which graphs have points of inflection | Every cubic has a point of inflection. |
| d^2y/dx^2 | Show change in concavity - useful when they want to know if inflection or not by going x = 5.1 and x = 4.9 to show the change about 5 If switches from +ve to -ve, is a point of inflection rather than max/min |
| Conditions for max point | dy/dx = 0 d^2y/dx^2 < 0 (or can be zero sometimes) |
| Conditions for min point | dy/dx = 0 d^2y/dx^2 > 0 (or can be zero sometimes) |
| Conditions for point of inflection | Can be 0 or not for dy/dx d^2y/dx^2 = 0 (when you get this =0, check both dy/dx both sides to see if actually inflection or not) |
| When d^2y/dx^2 switches from +ve to -ve | Not max/min - is a point of inflection. Can have stationary or non-stationary points of inflection (stationary means gradient = 0) |
| Chain rule | dy/dx = dy/du x du/dx Subbing in u for the contents of brackets e.g. (2x + 1)^10 du/dx means u differentiated with respect to x (the u isolated on one side) Remember to sub back in the u as the original always |
| If asked to find the other point of inflection | Still ALWAYS prove its a point of inflection |
| Rate always means | something/dt |
| e.g. radius r cm of circular ripple increasing at a rate of 8cms^1. At what rate is the area A increasing when radius is 25cm? | dA/dt (what we're looking for) = dr/dt (given in Q) x dA/dr (what the format must have via chain rule logic) A = pi*r^2 dA/dr = 2*r*pi Solve |
| How to get db/dx from a known value of dx/db | ^-1 |
| When you get an answer with multiple variables | Check in Q for a relationship between the two to get in terms of one variable. |
| e.g. sphere volume decreasing at rate of 60. Find rate of decrease of SA when volume is 800 | dV/dt = 60 V = 4/3*pi*r^2 dV/dr = 4*pi*r^2 A = 4*pi*r^2 dA/dr = 8*pi*r Looking for dA/dt dA/dt = dV/dt * dr/dV * dA/dr (look for nice diagonals cancelling) 800 = 4/3*pi*r^3 r = 5.8 Solve |
| Differentiated form of y = e^kx | dy/dx = ke^kx If the x is a function, do with chain rule |
| Differentiated form of lnx | 1/x Can only differentiate if a pure x is there - if it's a function, do with chain rule |
| Differentiate y = lnx^2 | =2lnx = 2/x Can do your normal log rearrangings first before differentiating |
| Product rule | If y = uv, then dy/dx = u*dv/dx + v*du/dx (differentiate each and multiply by the other one, then add) |
| Quotient rule | For when y = u/v In formula booklet |
| Removable solutions with logs | Negative logs - watch for them |
| Showing function is increasing | Must have line "f'x > 0" at some point |
| Vector notation | Underline them |
| Discriminant < 0 | So no *****REAL***** roots |
| Getting quick magnitude vector form | OPTN -> ANGLE -> POL(-3, 7) or OPTN -> ANGLE -> REC(sqrt58, 113 degrees) |
| Giving velocity answer | Give in VECTOR form. Works to do 4i + 10j - k |
| Easiest way to prove right angled | Pythagoras |
| Proving the 4th point (H) location on a parallelogram | Need to include: EF (with arrow above) = vector form EF = HG = vector GH = vector H = answer |
| If modulus after the flip is gradient = 3 | Gradient before the flip was -3 |
| Check for if =< is valid too | |
| Specify 'minimum point' for reasoning | |
| e.g. sec(2t) | MUST chain rule it |
| dy/dx(sinx) = | cosx |
| dy/dx(cosx) = | -sinx |
| dy/dx(tanx) = | sec^2(x) [can technically work out with quotient] |
| dy/dx(1) | 0 I'm not sure why this is here either honestly |
| For the denominator in the given equation | It uses the chance of it that variable in any branch |
| 'In context' | Say what sets A and B stand for |
| (-1)! | -1 * -2 * -3 * -4 ....... * -infinity |
| Doing choose for negative and fraction numbers | (-2)!/(3!)(-2-3)! Write out -2 * -3 * -4 etc for the negative ! on the top and bottom, going a few past where they cancel out and ending both on the same number for clarity. Then do ellipses. Then can cancel equal sequences |
| e.g. binomial expansion of 1/(1+x) | = (1+x)^-1 Write out for each expansion: -1C0 * 1 * x^0 = 1 * 1 * 1 = 1 -1C1 * 1 * x^1 -1C2 * 1 * x^2 etc Work out the chooses with the formula from before (ALWAYS WRITE IT) |
| Binomial expansion of e.g. (3 + 15x)^-4 | Can't do the weird binomial expansion unless one of the terms is 1. So do: [3(1 + 5x)]^-4 = 3^-4 * (1+5x)^-4 Also remember not just x for the expansion |
| 'Valid' expansions | i.e. converging function For e.g. (1 + 5x), the magnitude of 5x must be less than one for the value to converge to a value. Happens as x gets progressively smaller. Just set the magnitude of the not-1 part as less than 1, then solve for x |
| e.g. sqrt(1+x)/sqrt(1-x) | = (1+x)^0.5 * (1-x)^-0.5 Then binomially expand each separately to the amount of xs the Q asks = (binomial expansion 1) * (binomial expansion 2) = long list going up to x^4 roughly = short list only going up to x^2 (whatever asked in Q) |
| 'write the equations' of possible lines | Not x = 0, pi, 2*pi Must be: x = 0 x = pi x = 2pi |
| sec(2theta) + tan(2theta) | Remember can express tan as sin/cos to give common denominator |
| Partial fractions e.g. [5x]/[(x+4)(x-1)] | 5x/(x+4)(x-1) = A/(x+4) + B/(x-1) 5x = A(x-1) + B(x+4) Can sub in any value, so at x = -4, A(-5) = -20 etc |
| Partial fractions with more than 2 denoms e.g. [x+8]/[(x-2)(2x+1)(x+3)] | = A/(x-2) + B/(2x+1) + C/(x+3) x + 8 = A(2x+1)(x+3) + B...... etc Multiply each numerator by BOTH other denoms |
| Partial fractions with indices on the denoms e.g. (1+2x)/(x-1)^2 | Dont try to just split into equal denoms because then you can just add them. Instead so they have different denoms initially but can still be turned into the correct denom, do = A/(x-1) + B/(x-1)^2 A(x-1) + B = 1 + 2x |
| Partial fractions with indices on the denoms and multiple denoms e.g. (9x-14)/[(x+4)(x-1)^2] | = A/(x+4) + B/(x-1) + C/(x-1)^2 9x - 14 = A(x-1)^2 + B(x+4)(x-1) + C(x+4) to give everything the right denom they need weird multiplying |
| e.g. Express (2x^4 - 5x^3 - 11x^2 + 7x - 26)/[(x+2)(x-4)] in the form Ax^2 + Bx + C + D/(x+2) + E/(x-4) | The A,B,C show us we can divide Divide using quotient 2x^2 - x + 3. GIves remainder 5x -2 2x^2 - x + 3 + (5x-2)/(x^2 - 2x - 8) Partial fraction the end part. |
| Remainders (Y12 flashcard) | Remember: If the matrix method gives -15 when it should be -3, then the remainder is +12 (what you need to make it normal). |
| e.g. Find first 3 terms of binomial expansion of (7x-1)/(3-x)(1+3x) | Partial fractions give 2/(3-x) - 1/(1+3x) 2(3-x)^-1 = 2/3(1-(1/3)x)^-1 Then do the two binomial expansions and just subtract them |
| For chain rule of e.g. 1/5x | Dont just sub in u and then immediately unpack it treating it like an x. Do proper chain rule with dy/du |
| Implicit differentiation | Unlike explicit, doesnt have y as subject e.g. y^2 - 3xy = x^2 We must differentiate each term separately Remember constant terms *all* become zero |
| e.g. y^2 - 3xy = x^2 differentiated with respects to x - splitting into terms | d(y^2)/dx - d(3xy)/dx = d(x^2)/dx |
| e.g. y^2 - 3xy = x^2 differentiated with respects to x d(y^2)/dx = | d(y^2)/dy * dy/dx = 2y * dy/dx We cant differentiate y with respect to x - no sense. But we can do this by differentiating wrt y and then 'cancelling out' the dy |
| e.g. y^2 - 3xy = x^2 differentiated with respects to x d(3xy)/dx = | Product rule of 3x and y 3y + 3x * dy/dx [dy/dy just equals 1] |
| e.g. differentiate y = a^x | lny = lna^x lny = xlna 1/y * dy/dx = lna dy/dx = ylna y = a^x dy/dx = a^x*lna |
| Differentiation rule for a^kx | Original * loga * k |
| d(ln(6x))/dx how to calculate | ln6x = ln6 + ln(x) d(ln6x)/dx = 1/x |
| Need to memorise sum to infinity btw | |
| Parametric equations graphing | Sub in a few points with the form [theta = certain value]: (0,1) When making the table, make sure to use t to calculate y, not the x |
| Unit circle with parametrics | x = costheta, y = sintheta If the theta is 2theta instead, nothing changes: just doing same circle twice If 2costheta, same circle but radius 2 |
| Parametric equations | Variables expressed in terms of some independent parameter Not just combined by equating x in terms of y. Can also sub into other equation, add togehter, multiply, etc Trig identities useful Sub in to cancel fractions like 1/theta ^2 useful |
| e.g. x = 2t, y = t^2, -3<t<3 What is the domain/range of the function? | Highest/lowest values possible from the x/y. Make sure to put the equals to on the 0<= y, because unlike the other ones, can acc equal |
| Getting everything in terms of sin | Can do root version of 1 - cos^2 |
| x = t^2, y = t^3 | Remember negative. Calc won't show it |
| Differentiating parametrics | dy/dx = dy/dt x dt/dx So for parametrics: dy/dx = dy/dt divided by dx/dt |
| Gradient of normal | x to -1, not just reciprocal |
| 'has parameter 1/4 pi' | theta = 1/4 pi |
| If looking for exact value and costheta = -2/7 for a parametric | Keep it in terms of costheta and rephrase the variables |
| Often easier than weird integrations to calc area under curve | Integrate in terms of y. Rearrange into x = in terms of y, then definite integral as normal with the y values to go between |
| Area between two curves f(x) and g(x) | Given by top curve - bottom curve integral, while top curve ALWAYS >= bottom curve between a and b (use f[x] and g[x] for stating this) Negative areas don't matter - only calculating relative to each other |
| Area between curves bounded to y axis where they cross in the region | Would have to do two different integrals because they cross, so consider doing in terms of x instead to avoid this if relevant |
| Sideways quadratic curves where turning point of one is inside the other | The section above the turning point weird with dx, so do dy I don't really get this flashcard |
| integration of 1/-x | -lnIxI + c |
| integration of x^n | (1/[n+1])(x^n+1) |
| integration of e^x (opposite of dy/dx ones) | e^x + c |
| integration of 1/x (opposite of dy/dx ones) | (lnIxI) + c Remember modulus all the time Remember c for all integrals |
| integration of cosx (opposite of dy/dx ones) | sinx + c |
| integration of sinx (opposite of dy/dx ones) | -cosx + c |
| integration of sec^2(x) (opposite of dy/dx ones on formula sheet) | tanx + c |
| integration of cosec(x)cot(x) (opposite of dy/dx ones on formula sheet) | -cosec(x) + c |
| integration of cosec^2(x) (opposite of dy/dx ones on formula sheet) | -cotx + c |
| integration of sec(x)tan(x) (opposite of dy/dx ones on formula sheet) | secx + c |
| Integral notation | Always the brackets, always dx (or other variable) ALWAYS C |
| Integrating f(x+b), e.g. integral of (sqrt[x+2]) dx | integral of (sqrt[x+2]) dx = integral of (x+2)^1/2 = 2/3 (x+2)^3/2 + c Works exactly as you think it should |
| For any expansion where inner function is (ax + b) | Integrate as before and divide by a (including negative sign) |
| Integral of e^3x | 1/3 e^3x |
| Integral of (3x + 4)^3 | 1/12 (3x + 4)^4 |
| Don't raise power when integrating cos come on | |
| sqrt3/3 trying to get denominator of 9 | Don't x3: gives 3sqrt3. |
| Finding integrals with trig identities | Use all of them, including double angles (next few flashcards) |
| e.g. sin^2 integrated | use the cos2x one to sub in before the integration |
| e.g. sin3x * cos3x integrated | sin(2*theta) = 2sintheta*costheta theta = 3x 1/2sin6x = sin3x*cos3x Then integrate that |
| Show with addition formulae: sinAcosB = 1/2[sin(A+B) + sin(A-B)] | Expanding each of sin(A+B) and sin(A-B) cancels to 2sinA cosB Importantly, the formula in the Q of this flashcard is really useful for integration. e.g. sin3xcosx. |
| integral of 0.5(x+1) is mathematically equivalent to | 0.5* integral of x+1 |
| How to integrate by substitution [e.g. use substitution u = 2x+5 to find integral of x*sqrt(2x+5)] | Work out x (or sinx if easier) and dx in terms of u [x = (u-5)/2, du/dx = 2 so dx = 0.5 du]. Substitute in: 0.5du can halve integral instead (next card). Integrate the u Sub back in x |
| How the 0.5du thing works | Integrals have always been (stuff) * dx. Now it's just (stuff) * 0.5 * du so can isolate the du as needed |
| Which substitution to work out first for integrals | Work out the dx first to see if we actually need x (i.e. if the du coefficient will cancel the x). Still show moving the [e.g. 1/2x] into the brackets with the expression, then cancel with the x. Only rearrange to what you need e.g. 2x sub into 4x |
| Integration constants | Make sure all different. Combine in any other constants (i.e. even normal numbers) into a new constant at end. |
| Definite integrals when doing integration by substitution | If doing in terms of x, have to specify x = 0, x = 1 on each end of the integral. Better to work out new limits. Literally just u = 2x + 1 so sub in each limit |
| Integration by parts usage | Use when there are products to integrate. Formula in booklet. Essentially just treating one of the product terms as already being differentiated. Remember to add c |
| Integration by parts how to decide which is u and which is dv/dx | If there's a ln, do u = lnx (so you can easily differentiate it and get rid of the ln - must easier than integrating) If there's an x, do u = x so it becomes a 1 (lower priority than ln) |
| Estimates for integrals | Lower bound is where each bar starts |
| Show gradient of (expression) is always positive and deduce it has one real root only | For x much smaller than 0, y is negative (by subbing in). For very large x, y is positive. So there is at least 1 root and since there are no turning points there is only one root (doesnt double back on itself) |
| Show that the root of (strictly positive expression) lies between 0 and 1 | Sub in 0, sub in 1, show there's a change in sign so must be a root. 'As there is a change of sign, there must be a root between 0 and 1' - very explicit phrasing |
| Q rearranged equation 2x^3 + 4x - 5 = 0 to x = 5/(2x^2 + 4) Show its method | MUST show factorising: 2x^3 + 4x = 5 etc |
| Fixed point iteration e.g. looking for roots of equation 2x^3 + 4x - 5 = 0 Using iterative formula x[n+1] = 5/(2x[n]^2 + 4) and starting from x[0] = 1, find next two approxes x1 and x2 to the root | On sheet. Just subbing in x0, then x1, etc. Must show subbing in each value and the result. |
| Newton-Raphson method steps | Trying to find root of function f(x). We start with initial value x0. Take up to curve. Draw tangent. Where this meets x is the new x value. etc |
| Newton-Raphson method working for how we get the equation in the formula sheet (idk if we need this hopefully not) | At point (x0, f(x0)), tangent gradient is y - f(x0) = f'(x0)(x1-x0) When y = 0, -f(x0) = f'(x0)(x1 - x0) -f(x0)/f'(x0) = x1 - x0 x0 - f(x0)/f'(x0) = x1 0 and 1 to n and n+1 |
| sin3x differentiation | Must chain rule |
| Inflection | Concave upwards to concave downwards |
| Difference between integrals when one is negative | Make sure to subtract it still (so cancels into +) |
| Rate of cooling | Make sure negative |
| Differential equations: e.g. dv/dt = -3v^2 when t = 0, v = 10 Find v in terms of t | [v on one side with the dv/dt. else on other side] 1/v^2 * dv/dt = -3 [integrate both sides by dt. Cancel this dt to the /dt in the brackets so now wrt v on that side, *wrt t on the other*) *c1 and c2*. Combine constants to eg A. Sub in. Rearrange |
| 'Find general solutions to dy/dx = 2' | y = 2x + c |
| 'Find general solutions to dy/dx = -x/y' | Get the y to the left side so integrating the x will cancel Differential steps give y^2 = -x^2 + 2c y^2 + x^2 = 2c Let 2c = r^2 |
| What is a general solution | The constant of integration varies |
| How to find equation for lnIxI + lnIyI = c | To keep absolutes: lnIxyI = c IxyI = e^c = A xy = +-A y = +-A/x |
| Combining constants | Can even be like e^(kt+c) becoming Ae^kt That particular one is rlly necessary for future working |
| Trying to do differential equations for 1/y * dy/dx = 1/(x(1+x)) | Partial fraction it |
| Simplifying lnIyI = lnIxI - lnI1+xI + c | Make the c into lnA so you can combine all the lns into lnIAx/(1+x)I, then can remove the lns and moduluses |
| e.g. dy/dx = 2y + 3 | Needs dividing not subtracting, so divide by 2y + 3 |
| Subject of differential equation | Generally y, but trig function of y is fine |
| 'converse of theorem' | Swapping conclusion and requirement |
| 2lnIx+1I - lnIx-2I + c | Do the squared first before combining the two (so you dont square both) |
| Addition fomulae | Useful way to simplify stuff down, so look out for them. Make sure to also notice if e.g. A = x, B = 2x, as it is difficult to notice its an addition formula |
| Prove a number is divisible by 3 if and only if the sum of its digits is divisible by 3 and 9 | Pictures 16/04/26 |
| Let any 2 odd numbers be | Not both with n |
| Watch out sum vs product | |
| Prove if x^2 - px + q = 0 [p,q =/= 0] has 2 roots, one twice the other <=> 9q - 2p^2 = 0 | Pictures 16/04/26 |
| Prove that for a number >= 100 to be divisible by 4 it is necessary and sufficient that the number formed by its last 2 digits is a number divisible by 4 | Not a good proof in book - work out yourself |
| Proof by contradiction | Assuming the opposite of the proof then proving its impossible |
| Prove root 2 (and root 3 but same form) is irrational | Pictures 16/04/26 |
| Prove that there are infinitely many prime numbers | Pictures 16/04/26 |
| If Q mentions 3 consecutive positive integers | Then specify POSITIVE when defining n |
| Prove that the product of any 3 consecutive integers is even | Split into even and odd (useful to remember when even/odd buzzword used) Even: n = 2m so even product in the brackets Odd: n = 2m - 1. One of the brackets is (n+1) so cancels to 2m |
| Trapezium rule | h = b-a/n. n is the number of strips (so number of ordinates - 1) 1/2h (first y coord + last + 2(the rest)) Set up table of results to calc y: split x between upper and lower bound (inclusive) using h for counting up each, then work out y for each. |
| 'ordinate' | Y coordinate |
| Trapezium rule horrible es | Keep in terms of horrible es until the end |
| Trapezium rule checking | Use integral button on calc to check. May be very off if curve v steep (e.g. 500 rather than 800). Can be overestimate or underestimate |
| Iterative methods (i.e. fixed point iteration i think) | For when f(x) = 0. Rearrange into x = smth (can be xs) Draw curve g(x) and y = x Curve, line, repeat (alphabetical) Once decimals agree in a row, done |
| Terms for cobweb/spider when diverging | Same |
| How to tell if cobweb | Negative gradient at intersect with y = x Might staircase towards root then cobweb |
| Issues with fixed point iteration | Might not find all roots: Will only converge on a root with g(x) gradient at intersect of -1 to 1. Success varies on rearrangement |
| 'Given limit of sequence is k, show k satisfies ___' | 'limit' means when x[n+1] = x[n] = k Sub in for all of those |
| N-R how to choose starting point | Nearest point to the root. If between -2 and -1 given, choose either. Try both bounds for where value may lie to get expected answer. Or just make up a new point if neither bound locates the right root |
| All calculus angle type | Radians. That includes Newton-Raphson |
| 'To solve sinx = 0' | Go until root of pi found |
| Issues with newton raphson | If at stationary, impossible (will never touch x axis and is dividing by zero). If too close to stationary, will travel far before crossing axis so may be wrong root. |
| How does N-R work | Gradient of curve is y = f(x). Find gradient f'(x) and sub in from the starting x value. For each point, trace up to line, draw tangent down to cross axis, repeat. Sometimes diverges very far but comes back eventually. |
| How to use N-R to find stationary point (is usually for roots) | Make the dy/dx of the line the f(x) so its roots are when gradient = 0. Will need to differentiate that again for the formula |
| Differential equation expanding out brackets before isolating T | Consider not doing so - you can just divide by the brackets containing t |
| Remember absolute for integration | |
| Remember to write out finished expression, not just values of A and B | |
| 'Explain why this setup is satisfied by (formula)' | Define all terms of formula in words |
| 'Points on curve where gradient is 1 satisfy equation (stuff) = 0' | So the roots of the equation (stuff) = y are the points on the original curve where gradient 1 |
| 'Show how value in spreadsheet cell is calculated' | Say B3 = A3 * pi oe |
| 'In the expansion of (0.15 + 0.85)^50, the terms involving 0.15^r and 0.15^r+1 are denoted by...' | The terms include the 0.85 and constant. |
| 'Show his model fit the measurement' | Include conclusion |
| N-R labelling | Make sure to trace down each iteration to label x[1], x[2], etc |
| Proof pictures 09/05/26 | |
| How to differentiate e^(x^2) | For all e differentiation, the new coefficient is the differentiated form of the power So 2x * e^(x^2) |
| How to differentiate 2^(x^2) | Can do horrible substitution (earlier flashcard) Or just memorise f'(x) * ln(base) * f(x) i.e. 2x * ln2 * 2^(x^2) |
| Integrating x*e^(x^2) | Notice similar to known differential of 2x*e^(x^2) So integral(x*e^(x^2)) = 1/2 * e^(x^2) |
Created by:
Pyrogearos2