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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.
=(A6+B6)/2 =5*A6^3-7*A6+1 =5*B6^3-7*B6+1 =5*C6^3-7*C6+1 =IF(D6*F6<0,A6,C6) =IF(D6*F6<0,C6,B6) =(A7+B7)/2 =5*A7^3-7*A7+1 =5*B7^3-7*B7+1 =5*C7^3-7*C7+1 =IF(D7*F7<0,A7,C7) =IF(D7*F7<0,C7,B7) =(A8+B8)/2 =5*A8^3-7*A8+1 =5*B8^3-7*B8+1 =5*C8^3-7*C8+1 =IF(D8*F8<0,A8,C8) =IF(D8*F8<0,C8,B8) =(A9+B9)/2 =5*A9^3-7*A9+1 =5*B9^3-7*B9+1 =5*C9^3-7*C9+1 =IF(D9*F9<0,A9,C9) =IF(D9*F9<0,C9,B9) =(A10+B10)/2 =5*A10^3-7*A10+1 =5*B10^3-7*B10+1 =5*C10^3-7*C10+1 From this spread sheet I can find the upper and lower bounds of the intervals so that I can narrow down the accuracy of the root of the equation. Shown above are only the first few rows of the spreadsheet, to find the roots to a high enough accuracy, the formulas are filled down for more rows.
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this way only one of the two roots can be found, and thus resulting in a failure. The chosen formula is shown below: f(x) = -12.4x4-0.3x3+9.9x2-0.6x-0.6 From this it can be seen that the bisection will fail (from theory) however the proof and calculations are below: Lower Bound Middle Value Upper Bound Lower Bound Middle Value Upper Bound -1.000000 -0.500000 0.000000 x -1.000000 -0.750000 -0.500000 x -2.200000 1.437500 -0.600000 f(x) -2.200000 1.621875 1.437500 f(x) Lower Bound Middle Value Upper Bound Lower Bound Middle Value Upper Bound -1.000000 -0.875000 -0.750000 x -1.000000 -0.937500 -0.875000 x -2.200000 0.437012 1.621875 f(x)
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When using this method, you should start by taking increments in x of size 0.1 for the equation that we are using, assuming that by looking at the graph, you can see the two integers that a root is in between. For example: x f(x) 1 -2 1.1 -0.929 1.2 0.288 We then know that the root lies between 1.1 and 1.2 and we can say that the root is equal to 1.15 with a max error of �0.05. You could also say that the root is 0.1 to one decimal place.
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We can use this idea to show that: a1/2 = a a1/2 � a1/2 = a1/2 + 1/2 = a1 = a so: a1/2 = a In general a 1/n where n is a positive integer, is the nth root of a Division If the powers are being divided, indices of the same base can be subtracted. In general: an = an � am = an - m am a3 � a5 = a3 - 5 = a-2 Top There are two special cases: (a)
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1.88525391 1.88500977 -0.00077889 0.00024414 13 1.88500977 1.88525391 1.88513184 0.00037087 0.00012207 14 1.88500977 1.88513184 1.88507080 -0.00020399 0.00006104 15 1.88507080 1.88513184 1.88510132 0.00008345 0.00003052 16 1.88507080 1.88510132 1.88508606 -0.00006027 0.00001526 17 1.88508606 1.88510132 1.88509369 0.00001159 0.00000763 18 1.88508606 1.88509369 1.88508987 -0.00002434 0.00000381 The formulae I used in the spreadsheet are shown below: a b (a+b)/2 f(x) error 1 1 2 =1/2*(B2+C2) =(D2-2)*(D2-4)*(D2-6)+1 =1/2*(C2-B2) 2 =IF(E2<0,D2,B2) =IF(E2<0,C2,D2) =1/2*(B3+C3) =(D3-2)*(D3-4)*(D3-6)+1 =1/2*(C3-B3) 3 =IF(E3<0,D3,B3) =IF(E3<0,C3,D3) =1/2*(B4+C4) =(D4-2)*(D4-4)*(D4-6)+1 =1/2*(C4-B4) 4 =IF(E4<0,D4,B4) =IF(E4<0,C4,D4) =1/2*(B5+C5) =(D5-2)*(D5-4)*(D5-6)+1 =1/2*(C5-B5) 5 =IF(E5<0,D5,B5) =IF(E5<0,C5,D5) =1/2*(B6+C6) =(D6-2)*(D6-4)*(D6-6)+1 =1/2*(C6-B6) 6 =IF(E6<0,D6,B6) =IF(E6<0,C6,D6) =1/2*(B7+C7) =(D7-2)*(D7-4)*(D7-6)+1 =1/2*(C7-B7) 7 =IF(E7<0,D7,B7) =IF(E7<0,C7,D7) =1/2*(B8+C8) =(D8-2)*(D8-4)*(D8-6)+1 =1/2*(C8-B8) 8 =IF(E8<0,D8,B8) =IF(E8<0,C8,D8)
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f(x) = x� - 12x + 5 I am going to use the root in the interval [0,1] f(0) > 0 f(1) < 0 f(0+1) � f(0.5) = -0.875 2 Therefore f(0.5) < 0 f(0) > 0 f(0.5) < 0 f(1) < 0 Root is in interval [0, 0.5] I now test the mid-point of [0, 0.5] f(0+0.5) � f(0.25) = 2.015625 2 Therefore f(0.25) > 0 f(0) > 0 f(0.25) > 0 f(0.5) < 0 Root is in interval [0.25, 0.5] I will now test mid-point of [0.25, 0.5] f(0.25+0.5) � f(0.375) = 0.552734375 2 Therefore f(0.375) > 0 f(0.25)
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EQUATION USED: = 0 has only 1 roots in [0,-1] From Graphmatica: Results obtained from Microsoft Excel: X-value F(X) -1 -1.5 0 0.5 Observe the change of y-value from negative value to positive value in the table. That shows that the root of the equation where the y-value is 0, is between x = 0 and x = -1. Now this is step 1, we're moving on to step 2. X-value F(X) -0.9 -0.7882969 -0.8 -0.3497152 -0.7 -0.0723543 -0.6 0.1120064 Now the search is narrowing in and it comes to one decimal place in between x = -1 and x = 0.
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We know from the rules of differentiation for such a function that for the (n+1) th differential you multiply the nth function by its power and lower its power by one. However, if we don't know the nth differential and we know only the original function this simple method is obviously not possible. Looking at the above example suggests that a solution would have to have something to do with factorials. The way the power and the differential number add together to equal b is of particular interest.
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Fractals. In order to create a fractal, you will need to be acquainted with complex numbers. Complex numbers on a graph are characterized by the coordinates of (x,y)
An example of a complex number can be: (5 + 4i). Complex numbers look ambitious, but they are not. In order to add complex numbers, you just need to add like terms:. In order to multiply complex numbers, you need to use the distributive law: . If you are faced with i2, please note that it is equal to -1! We shall start with Mandelbrot?s set. The Mandelbrot set was popularized due to its aesthetic appeal to many, and the simple rules that were applied in order to generate a complex structure.
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