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# AS and A Level: Core & Pure Mathematics

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## Differentiation and intergration

1. 1 It is easy to get differentiation and integration the wrong way round. Remember that the power gets smaller when differentiating.
2. 2 Differentiation allows you to find the gradient of a tangent at any point on a curve. The first derivative describes the rate of change.
3. 3 If a function is increasing then the first derivative is positive, if a function is decreasing, then the first derivative is negative.
4. 4 When asked to find the area under a curve, it is asking you to integrate that curve between two points. Even if you don’t know the points, pick two numbers. You’ll get marks for methods.
5. 5 When referring to a min/max/stationary point, the gradient equals 0. Differentiate the curve and set this to equal 0. The second derivative tells you whether it is a maximum or minimum. If the second derivative is positive, the point is a minimum, if the second derivative is negative, then the point is a maximum.

1. 1 When solving a quadratic inequality, always draw a picture. The inequality is less than 0, where the curve is below the x-axis and bigger than 0 when the curve is above the x-axis.
2. 2 Sometimes in part (a) of a question you are asked to find something, for example a radius. In part (b) you might be then asked to use the radius that you found. If you couldn’t do part (a), don’t give up, choose a random radius.

## Straight lines

1. 1 To find the distance of a straight line, draw the straight line with the co-ordinates. Then make a right angle triangle, find the lengths of the horizontal and vertical lines, then use Pythagoras.
2. 2 When a question asks you for a straight line. The first thing to do is to write down the equation of a straight line. Then find out what information you know, and what information you need. Even if you don’t understand the whole question, it is important to start.

• Marked by Teachers essays 3
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1. ## C3 Coursework - different methods of solving equations.

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I repeat this until I get down to increments the size of 0.00001 x F(x) -3.1 8.429 -3.2 6.712 -3.3 4.843 -3.4 2.816 -3.5 0.625 -3.6 -1.736 -3.7 -4.273 -3.8 -6.992 -3.9 -9.899 -4.0 -13 x F(x) -3.51 0.396649 -3.52 0.166592 -3.53 -0.06518 -3.54 -0.29866 -3.55 -0.53387 -3.56 -0.77082 -3.57 -1.00949 -3.58 -1.24991 -3.59 -1.49208 -3.60 -1.736 x F(x) -3.521 0.143492 -3.522 0.120375 -3.523 0.097241 -3.524 0.07409 -3.525 0.050922 -3.526 0.027736 -3.527 0.004534 -3.528 -0.01869 -3.529 -0.04192 -3.530 -0.06518 x F(x) -3.5271 0.002213 -3.5272 -0.00011 -3.5273 -0.00243 -3.5274 -0.00475 -3.5275 -0.00707 -3.5276 -0.0094 -3.5277 -0.01172 -3.5278 -0.01404 -3.5279 -0.01636 -3.5280 -0.01869 x F(x)

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This is the formula used for this method: Gradient = Vertical (change in y axis) Horizontal (change in x axis) Therefore the gradient of the curve equals - Gradient = 12-4/4-2 = 8/2=4 This is most likely wrong and most certainly not accurate - it is only estimation. The second method, the increment method is far more accurate. Increment method - This is where you plot two points, represented by the letters P and Q, on the curve, and draw a line to join them.

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3. ## Estimate a consumption function for the UK economy explaining the economic theory and statistical techniques you have used.

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That is...dC/dY is positive and less than unity."(Keynes, 1936,p.96) From this statement can be derived the Keynesian consumption function (Absolute income) which is usually expressed in the following way: Ct=c0+c1Yt Where Ct is consumption, c0 is autonomous consumption, Yt is income and c1 is the marginal propensity to consume. The Keynesian view is that when income rises, consumption will rises as well, but less than income. This indicates that c1, the marginal propensity of consume, is lees than l. Keynes believed that the main influence on consumption is consumers' disposable income. When their disposable income fluctuates, the consumption would follow.

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4. ## Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and also disadvantage in these methods and they will be discussing throughout this investigation.

on both sides of the root. The mid point, c is then found by formula. Then f(c) is calculate, the purpose of calculate f(c) is to define the position on c. If f(c) > 0 then it means the range does not contain the roots and there fore the roots must between. We will then repeat the same process and reduce the range of root until the data approach to the solution, although we can never get the exact value of the root but we will be able to approach to the root to 5 decimal places (5d.p).

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5. ## I am going to solve equations by using three different numerical methods in this coursework and compare the methods which are Bisection, Newton-Raphson, and Rearranging method.

I am going to combine the graph and the figure I have worked out in Excel. As the figures show blow: A B f(a)<0 f(b)>0 (a+b)/2 y=2x�-3x�-8x+7 -2 -1 -5 10 -1.5 5.5 -2 -1.5 -5 5.5 -1.75 1.09375 -2 -1.75 -5 1.09375 -1.875 -1.73047 -1.875 -1.75 -1.73046875 1.09375 -1.8125 -0.26416 -1.8125 -1.75 -0.264160156 1.09375 -1.78125 0.428162 -1.8125 -1.78125 -0.264160156 0.428162 -1.796875 0.085365 -1.8125 -1.796875 -0.264160156 0.085365 -1.8046875 -0.08855 -1.804688 -1.796875 -0.088553429 0.085365 -1.80078125 -0.00138 -1.800781 -1.796875 -0.001383424 0.085365 -1.798828125 0.042044 -1.800781 -1.798828125 -0.001383424 0.042044 -1.799804688 0.020343 -1.800781 -1.799804688 -0.001383424 0.020343 -1.800292969 0.009483 -1.800781 -1.800292969 -0.001383424 0.009483 -1.800537109 0.004051 -1.800781 -1.800537109 -0.001383424 0.004051 -1.80065918 0.001334 -1.800781 -1.80065918 -0.001383424 0.001334 -1.800720215 -2.5E-05 -1.80072 -1.80065918 -2.47371E-05 0.001334 -1.800689697

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6. ## Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson method

=0, the values of x for which the graph of y=f(x) crosses the x axis is expected to achieve. A characteristic is also needed to be mentioned is that when the curve crossed the x axis, which means y =0, the f (x) sign change, it indicate that interval must contain a root. After we spot a root in the interval, the mid-point of the interval will be taken. Graph 1.1 ---y = (2x-3) (x+1) (x-2)-1 When the graph of y = f(x) cross the x axis, f(x) change the sign of its value. To solve the equation by bisection iteration, the values of f(x)

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7. ## Different methods of solving equations compared. From the Excel tables of each method, we know that method 1 (change of sign method) takes 28 steps to find the root, while method 2 (Newton-Raphson method) and method 3 (rearrangement) take 4 and 17 steps r

The table below shows the data with accuracy of 8 decimal places. a f(a) b f(b) c f( c) Max. error 0.00000000 2.60000000 1.00000000 -0.80000000 0.50000000 1.00000000 1.70000000 0.50000000 1.00000000 1.00000000 -0.80000000 0.75000000 0.10625000 0.90000000 0.75000000 0.10625000 1.00000000 -0.80000000 0.87500000 -0.34765625 0.45312500 0.75000000 0.10625000 0.87500000 -0.34765625 0.81250000 -0.12060547 0.22695313 0.75000000 0.10625000 0.81250000 -0.12060547 0.78125000 -0.00711670 0.11342773 0.75000000 0.10625000 0.78125000 -0.00711670 0.76562500 0.04958649 0.05668335 0.76562500 0.04958649 0.78125000 -0.00711670 0.77343750 0.02123928 0.02835159 0.77343750 0.02123928 0.78125000 -0.00711670 0.77734375 0.00706232 0.01417799 0.77734375 0.00706232 0.78125000 -0.00711670 0.77929688 -0.00002694 0.00708951 0.77734375 0.00706232 0.77929688 -0.00002694 0.77832031 0.00351775 0.00354463 0.77832031 0.00351775 0.77929688 -0.00002694 0.77880859 0.00174542 0.00177235 0.77880859 0.00174542 0.77929688 -0.00002694 0.77905273 0.00085924 0.00088618 0.77905273 0.00085924 0.77929688 -0.00002694 0.77917480 0.00041615 0.00044309 0.77917480 0.00041615

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8. ## Numerical integration can be described as set of algorithms for calculating the numerical value of a definite integral. Definite integrals arise in many different areas and calculus is a tool

The point at which the dashed line meets the curve is the height (expressed as ??in algebraic terms?. From basic arithmetic given the height and width we can calculate the area (?h). By calculating the area of each of the rectangles in figure 1.0 and adding the results we obtain the approximation to the area under the curve (between 0 and 1 radians). Through the use of algebra it is possible to derive the general from of the mid-point rule, using n, each of width h. The approximation of can be given by: > the mid-points of the five rectangles are ????????h/2 ??= ?????h/2 ??= ?????h/2 ?????????h/2 ?????????h?? > the heights of these rectangles are respectively f(??) f(???

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Then the final step, how to find the gradient using the differences. To find the gradient, you have to use the method: GRADIENT = VERTICAL DIFFERENCE HORIZONTAL DIFFERENCE GRADIENT= 9 1.5 = 6 So the gradient at x=3 is 6. This shows an example of how to obtain the gradient at different "x" values. In my investigation, I will first start off by investigating simple graphs, such as straight line graphs. Then I will investigate further onto "Parabola" curves. Straight line graphs are simple and comprehensible. Parabola curves consist of a power which is larger than 1.

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10. ## In this investigation I am going to investigate three methods of finding the roots to equations and then compare them. I will be using those techniques fully. The three methods that I am going to examine are:

The most ingenuous way to do this is to search for where the value of f(x) changes. This, of course, will only occur when the line crosses the x-axis locating the roots. Because the number line is infinite we cannot just check all the integer values from -? to ? so it is always a good idea to plot a graph of the equation to find out what region its roots lie in. The equation that I have chosen to demonstrate how Decimal Search is successful is f(x)=3x3-7x2-11x+17. In order to find the root of the function f(x)=3x3-7x2-11x+17, the decimal search method was used.

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11. ## Functions Coursework - A2 Maths

x f(x) -5 -111 -4 -53 -3 -19 -2 -3 -1 1 0 -1 1 -3 2 1 3 17 4 51 5 109 One can see that there is a change of sign, and therefore a root of f(x)=0, between x=1 and x=2. If I wanted to find the root of the equation between x=1 and x=2 to five decimal places, I would do the following: The change of sign method will be applied for values of x between 1 and 2, taking 0.1 to be the interval in x values.

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12. ## Solving Equations Using Numerical Methods

Here is the table of results. From this table, I can see that the change in sign appears in between -1.53 and -1.52. To be certain that the root is between -1.53 and -1.52, I will calculate f(x) in each case to make sure that a change of sign occurs. F(x) = y=x�-2x+0.5 F(-1. 53) = -0.0216 F(-1. 52) = 0.0282 As a change of sign has been found and confirmed here, the root does lie between these points. I will now use the same method and use increment of 0.001 to find the root between -1.53 and -1.52.

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13. ## MEI numerical Methods

Then test this value to see if it's positive or negative, dependent on the result the interval then becomes smaller, hence the answer gets closer to the real answer (converges). This method can be used to solve this equation as we have been given the interval estimate, 0 < x < ?/2. Hence to obtain a better estimate we test, (0 + ?/2) / 2, which is ?/4. We now do f(?/4) in order to see if its negative or positive hence making the interval smaller.

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14. ## Mathematics Coursework - OCR A Level

I then chose to look at the x values between these two values. I decreased the x value by 0.001 between x = -1.07 and -1.08. The change of sign between these two values occurred between x = -1.073 and x = -1.074. The root needs to be given to 3 decimal places and therefore I need to check the values between these two values to see whether the root is closer to the lower or higher value of x. To 3 decimal places, one root of the equation y=0.5x5-3x3-3 is x = -1.074 as the change of sign occurs between the interval x= -1.0737 and x = -1.0738 therefore I have taken

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15. ## Arctic Research (Maths Coursework)

On the other hand, if it was considered the flight of the plane would be affected. * The flight velocity of the plane is constant and the speed at which the plane travels will be achieved instantaneously. This will prevent complications concerned with takeoff and landing times. * Wind velocity is also constant and always blowing in the exact, same direction of the west, as the flight times for a constantly changing wind velocity would be almost impossible to calculate.

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16. ## Number pyramids

In that case, my formula is as follows: - (F + L) x 2 = T By using this formula I can calculate any first, last and top number. Test my formula (F + L) x 2 = T The number pyramid I am going to test has base numbers of 10, 11 and 12. Using this I will work out the top number. Formula: (F + L) x 2 = T Substituting: (10 + 12) x 2 = 44 By substituting the formula into the number pyramid, you will notice that: - 44 21 23 10 11 12 By

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17. ## Math Portfolio Type II - Applications of Sinusoidal Functions

June 1 152 4.60h July 1 182 4.62h August 1 213 5.07h September 1 244 5.65h October 1 274 6.22h November 1 305 6.87h December 1 335 7.50h All the values of the sunrise times for Toronto are converted to a decimal. The values of the sunrise times (y-value on the graph) and the number of days (x-value on the graph) are listed into Microsoft Excel. A graph is created from the values and everything on the graph is labelled.

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18. ## Estimate a consumption function for the UK economy explaining the economic theory and statistical techniques you have used.

difference between our two lines and as you can see, the equation holds well until the late 1970's where large negative differences start to appear. This is because during this period there was the oil crisis and high unemployment and so there was a lot of uncertainty in the economy. This meant that people were unsure of their future income and so consumed less. Here we have the first failing of the Keynesian consumption function, people decide how much they consume not only on what they are earning now but also what they expect to be earning in the future.

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19. ## Solutions of equations

I will need to investigate further and check for roots between 0 and 1. The table of values is given below. Table 2 x F(x) 0 -32 0.1 -16.33 0.2 -6.776 0.3 -1.859 0.4 -0.128 0.5 -0.125 0.6 -0.392 0.7 0.529 0.8 4.096 0.9 11.767 1 25 So I know there is a root between [0.6,0.7]. This only accounts for root B. What about Root A? Here is an example of the method failing. I will go into this in more detail further on. Root B, However still needs to be found more accurately.

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20. ## The open box problem

If we then multiply these sides together and then by x to get the volume we end up with V= x(6-2x)^2. I will now construct a table to show a range of values for x and the volume of the open box, using the equation. X 0.5 1.0 1.5 2.0 2.5 3 V (volume) 12.5 16 13.5 8 2.5 0 We can see here that the maximum volume for the dimensions 6x6 lies between 0.5 and 1.5; so I will now draw another table that focuses in more between the numbers 0.5 and 1.5.

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21. ## Triminoes Investigation

The game is played using triangular pieces of card. Each card has 3 numbers on it. I have to investigate the relationship between the number of trimino cards in a set and the largest number on the cards and I also have to find the relationship between the sums of all cards in a set of Trimino cards and the largest number used on the card. To find these relationships, I have to do the basics, which are: To find: 1. The number of Triminoes cards 2. Largest number used on the cards 3. Sum of all the numbers This diagram shows the set of ten Triminoes cards used for a game.

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22. ## Examining, analysing and comparing three different ways in which to find the roots to an equation.

To investigate these values you put them into the formula to receive a value of f(x), and you can thus see where the change of sign occurs. When this has been found, it will be again between two values, so again you adjust the decimal places by one, and you are now investigating to 2 decimal places, such as 3.22 and 3.23. You keep going further into more decimal places, in order to increase the accuracy of root. For my investigation I will be going to 3 decimal places. To investigate this method I will be using this equation: f(x)

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23. ## Analyse the use of three methods which are called the: change of sign, Newton-Raphson and the rearrangement method and use them to find roots of different equations.

The equation I am going to find the root of is f(X)= x5+4x4-6x which looks like this. As shown above there are 3 roots one which is at 0 and the others between -5 and -4, 1 and 2. I'm going to find the root which lies between 1 and 2. To find this root I'm going to use excel to see where the solution of the equation changes sign. This shows the first few columns to how we came to find the root between 1 and 2.

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24. ## Solution of equations by numerical methods.

at the point (x1, f(x1)). The point at which the tangent cuts the x-axis gives the second approximation for the root, and so the process is repeated. The gradient of the tangent at (x1, f(x1)) is f'(x1), and as the equation of a straight line can be written y - y1 = m(x - x1), the equation of the tangent can be said to be y - f(x1) = f'(x1)[x - x1]. This tangent cuts the x -axis at (x2, 0) so we can say that 0 - f(x1)

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25. ## Change of Sign Method.

I will continue the process using the increments of 0.001 within the interval [0.83,0.84]. x 0.830 0.831 0.832 0.833 0.834 0.835 f(x) -0.044089 -0.039211927 -0.034320896 -0.029415889 -0.024496888 -0.019563875 x 0.836 0.837 0.838 0.839 f(x) -0.014616832 -0.009655741 -0.004680584 0.000308657 The root therefore lies in the interval [0.838,0.839]. I shall now use increments of 0.0001 in the interval [0.838,0.839]. x 0.8380 0.8381 0.8382 0.8383 0.8384 0.8385 f(x) -0.004680584 -0.00418229398 -0.0036838631 -0.00318529134 -0.00268657869 -0.00218772512 x 0.8386 0.8387 0.8388 0.8389 0.8390 f(x) -0.00168873063 -0.00118959519 -0.00069031878 -0.00019090139 0.000308657 Therefore the root lies between 0.8389 and 0.8390. To three decimal places, the root = 0.839.

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