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AS and A Level: Core & Pure Mathematics
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Differentiation and intergration
- 1 It is easy to get differentiation and integration the wrong way round. Remember that the power gets smaller when differentiating.
- 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 If a function is increasing then the first derivative is positive, if a function is decreasing, then the first derivative is negative.
- 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 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.
Quadratics and circles
- 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 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.
- 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 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
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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Estimate a consumption function for the UK economy explaining the economic theory and statistical techniques you have used.3 star(s)
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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1.2991 -0.000805 1.2992 0.000720 1.2993 0.002244 The root of this equation lies in the interval [1.2991, 1.2992]. This means that I can take the root to be x = 1.29915 (5 decimal places). The maximum error of the root would be 0.00005 However, there are some functions for which the Decimal Search method would not work. An example is (1.55x+3.4)4=0. This would not work because the curve is touching the x-axis, so there would not be a change of sign.
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Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically
The error bounds are 0.3095 and 0.310. Change of Sign Method Failing Consider the equation f(x) =0, where f(x) = I shall try to locate the roots by making a table of values that correspond to this function of x. Rearranging f(x) =0 into x=g(x) Rearranging f(x) =0 into x=g(x) failing I am going to solve the equation f(x) =0 where f(x) =. The graph of y=f(x) is shown below. ==> ==> ==> ==> ==> ==> The sequence formed is converging; however it converges to the incorrect root meaning that it is considered as a failure.
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Now use decimal search and Excel within the interval [2.560, 2.570] x f(x) 2.561 -0.00356 2.562 0.00288 2.563 0.00933 There is a root in the range (2.561, 2.562) Now use decimal search and Excel within the interval [2.5610, 2.5620] 2.561 -0.00356 2.5611 -0.00291 2.5612 -0.00227 2.5613 -0.00163 2.5614 -0.00098 2.5615 -0.00034 2.5616 0.000304 There is a root in the range (2.5615, 2.5616) Now use decimal search and Excel within the interval [2.5615, 2.5616] x f(x) 2.56150 -0.00034 2.56151 -0.00028 2.56152 -0.00021 2.56153 -0.00015 2.56154 -8.2E-05 2.56155 -1.8E-05 2.56156 4.63E-05 There is a root in the range (2.56155, 2.56156)
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Exponential decay:The objective of this experiment is to get the trend line of heart rate and to determine the decay constant of this decay.
2.1 Apparatus 17 people, stop watch, playground. 2.2 Procedures 1. Everyone put their hands on carotid artery. 2. Timekeeper calculate the time use. 3. Once the timekeeper call "start!", students can begin to count the heartbeat. 4. Repeat 3 times and get 3 different groups of data. 5. All the students run 5 minutes continuously. 6. Repeat the procedure 1-3 every half a minutes for 7 times. 2.3 Data To get a obvious and simple curve, for each item, I just calculate the average value.
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In my coursework I will be using three equations to investigate their solutions using three numerical methods which are: change of sign using Newton-Raphson by finding the fixed point iteration fixed point iteration after rearranging the equa
The next value I'll try is the midpoint of the new interval, -0.75 f(-0.75) = 0.34375 f(-0.875) = -0.66797 So I can see that the root lie between -0.75 and -0.875 So I now try and find the root by finding f(x) of the mid point between [-0.75, -0.875] f(-0.8125) = -0.12354 f(-0.78125) = 0.119568 to do this method I have used a calculator f(-0.79688) = 0.000365 and recorded my values in a table on excel Decimal Search A follow on from the bisection I can see that the root is between [-0.8125, -0.79688] so for this method I will try the different decimals until I become 7 d.p from the root.
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For example if the equation has a repeated root as shown below for the equation x�-0.96x�-5.6471x+7.48225=0. As the root between 1 and 2 is repeated, the line of the graph never crosses the x axis so the function does not ever take a negative value between 1 and 2. As a result of this no change of sign will be found using the method above and so the root will not be found. Newton-Raphson Method The Newton-Raphson method takes a value of x close to the root which is to be found and then uses and iterative formula to generate a value which is closer to the root.
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Known: Segment AC=1 Segment BC=1 Angle ACD=60 We know two sides and one included angle. Then we could use the formula to find the area of triangle. As we known, the area of triangle ACD is one sixth of area of hexagon. Let us try 12-sided, 24-sided, and 48-sided polygons. 12-sided: The triangle ACD in 12-sided is one twelfth of the . Known: Segment AC=1 Segment BC=1 Angle ACD=30 Area of triangle ACD: Area of : 24-sided: The triangle ACD in 24-sided is one twenty-fourth of the . Known: Segment AC=1 Segment BC=1 Angle ACD=15 Area of triangle ACD: Area of : 48-sided: The triangle ACD in 48-sided is one forty-eighth of the .
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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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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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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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where h = (w - a) cos (?) (height) a = a (first parallel side) b = a + 2(w - a) sin (?) (second parallel side) To calculate angle ?, I have taken the vertical perpendicular to the base as the starting point at ? = 0, Firstly, due to the 3 variables I fixed a and w-a to be equal length and varied ? by 1 degree. A snapshot of the values obtained using excel is below. ?� with horizontal a b Area 28 3.333 3.333 14.41629 29 3.333 3.333 14.42938 30 3.333 3.333 14.43376 31 3.333 3.333 14.42935 32 3.333 3.333 14.41606 Highlighted in red using conditional formatting is the greatest area using these values and varying ?
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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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The method I am going to use to solve x−3x-1=0 is the Change Of Sign Method involving the Decimal Search method
Therefore, I can narrow there values down further to find another change of sign. x f(x) -0.31 -0.09979 -0.32 -0.07277 -0.33 -0.04594 -0.34 -0.01930 -0.35 0.00713 A I can see there is a change of sign between x=-0.34 and x=-0.35. Therefore, I can narrow there values down further again to find another change of sign. x f(x) -0.349 4.491451 x 10^-3 -0.348 1.855808 x 10^-3 -0.347 -7.81923 x 10^-4 I can see there is a change of sign between x=-0.347 and x=-0.348.
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This is because they are so close together that there is no sign change between f(0) and f(0.1). The graph on the next page shows the graph of y=f(x), which shows that the roots cannot be found using a decimal search. Newton-Raphson Finding a root The graph of y=f(x): Roots of f(x)=0 exist in the intervals [-2,-1] and [0,1]. f(x)=(x2+0.9x-2.52)-1+1 f'(x)=-(2x+0.9)(x2+0.9x-2.52)-2 We shall find a root by taking x1=1 xn f(xn) f'(xn) x1 1 -0.61290 -7.54422 x2 0.91876 -0.17786 -3.79788 x3 0.87193 -0.02563 -2.78114 x4 0.86271 -0.00071 -2.62915 x5 0.86244 0.00000 -2.62488 x6 0.86244 0.00000 -2.62488 f(0.862435)=0.000014 f(0.862445)=-0.000012 The change of sign indicates that the root is 0.86244�0.000005.
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Functions. Mappings transform one set of numbers into another set of numbers. We could display a mapping in three ways
Inverse Functions If a function is one-to-one then an inverse function exists, called f-1(x) For Simple Functions eg f(x) = 4x + 1, x?R x Alternatively, swap y and x throughout and rearrange to make y the subject. Many-to-one functions can be made into one-to-one functions by restricting the domain. f-1(x) can then be found Range of function = Domain of Inverse Function Domain of Function = Range of Inverse Function Sketching the graph can be helpful to determine the range and the domain. y = f(x) is a reflection of y = f-1(x) in the line y = x An asymptote is formed where there is a "forbidden" number.
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The approximate methods of definite integrals may be determined by numerical integration using: 1. The Trapezium Rule: The Trapezium Rule divides the area underneath the curve into trapeziums. We can then use the formula (where a, b are the bases and h is the height of the trapezium) to estimate the area. Dependent on the amount of trapezia (n) used the general formula is: 2. The Midpoint Rule: The Midpoint Rule divides the area underneath the curve into rectangles. We can then use the formula (where a, b are the 2 different sides of the rectangle)
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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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0.7] [0.66, 0.67] [0.662, 0.663] Root intervals [0.6621, 0.6622] I know that the root is 0.662 to 3 decimal places In order to test the boundaries, I put them back into the equation: (0.6621)4 - 5(0.6621)2 +2 = 0.00292 (0.6622)4 - 5(0.6622)2 + 2 = -0.0025 As the values have different signs, it further indicates that the root is close to 0.662 to 3 decimal places. Failure The decimal search will generally fail when two roots are close together or when there is a double root present.
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Find a general pattern for log x + log y Log x + log y = log (xy) (e) Can you suggest why this is true? This rule is true because when you multiply both of the abscissas they equal another abscissa and adding the two logs will equal the third log. Therefore the general pattern is true. 2. (a) Copy and complete the following table using your calculator. Give your answers to the correct four decimal places. Log 12 - log 3 0.6020 Log 4 0.6020 Log 50 - log 2 1.3979 Log 25 1.3979 Log 7 - log
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I will now illustrate one of these roots graphically, showing closely the changes in the tangent to the curve. I will use the root in interval [2, 3] Error bounds for interval [2, 3] 2.3301 is the root in this interval to 5 significant figures. Therefore the error is 2.3301 � 0.00005 I will now perform the change of sign test to confirm it is within these limits. Lower limit is 2.33005 then f (2.33005) = -0.000098648 Upper limit is 2.33015 then f (2.33015) = 0.0010302 There is a change of sign which confirms root in interval is 2.3301 � 0.00005 When does this method fail?
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X Y -0.3 0.179 -0.29 0.133233 -0.28 0.087744 -0.27 0.042551 -0.26 -0.00233 -0.25 -0.04688 -0.24 -0.09107 -0.23 -0.1349 -0.22 -0.17834 -0.21 -0.22138 -0.2 -0.264 There is a change of sign and therefore a root in the region between -0.27 and -0.26. The next part of this investigation is finding where, between -0.27 and -0.26, a change of sign occurs. X Y -0.27 0.042551 -0.269 0.038049 -0.268 0.03355 -0.267 0.029054 -0.266 0.024561 -0.265 0.020071 -0.264 0.015585 -0.263 0.011102 -0.262 0.006622 -0.261 0.002145 -0.26 -0.00233 There is a change of sign and therefore a root in the region between -0.261 and -0.260.
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This table shows the results of the numbers in increments of 0.1 between 0 and 1. Upon looking at the results we can see that there is a change of sign between 0.7 and 0.8, this shown by the graph below. This method is repeated again with intervals of 0.01 between 0.7 and 0.8 The table shows that there is a change if sign between 0.75 and 1.76. This means that the root of the equation lies between 0.75 and 0.76. This is shown by the graph below. Once again this method is repeated with intervals of 0.001 between 0.75 and 0.76.
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