AS and A Level: Core & Pure Mathematics
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Change of sign method.
=(A6+B6)/2 =5*A6^37*A6+1 =5*B6^37*B6+1 =5*C6^37*C6+1 =IF(D6*F6<0,A6,C6) =IF(D6*F6<0,C6,B6) =(A7+B7)/2 =5*A7^37*A7+1 =5*B7^37*B7+1 =5*C7^37*C7+1 =IF(D7*F7<0,A7,C7) =IF(D7*F7<0,C7,B7) =(A8+B8)/2 =5*A8^37*A8+1 =5*B8^37*B8+1 =5*C8^37*C8+1 =IF(D8*F8<0,A8,C8) =IF(D8*F8<0,C8,B8) =(A9+B9)/2 =5*A9^37*A9+1 =5*B9^37*B9+1 =5*C9^37*C9+1 =IF(D9*F9<0,A9,C9) =IF(D9*F9<0,C9,B9) =(A10+B10)/2 =5*A10^37*A10+1 =5*B10^37*B10+1 =5*C10^37*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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Change of sign.
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) = 12.4x40.3x3+9.9x20.6x0.6 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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Solving Equations Numerically
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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Looking at Indices
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 = a2 Top There are two special cases: (a)
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Change of sign method  interval bisection method
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) =(D22)*(D24)*(D26)+1 =1/2*(C2B2) 2 =IF(E2<0,D2,B2) =IF(E2<0,C2,D2) =1/2*(B3+C3) =(D32)*(D34)*(D36)+1 =1/2*(C3B3) 3 =IF(E3<0,D3,B3) =IF(E3<0,C3,D3) =1/2*(B4+C4) =(D42)*(D44)*(D46)+1 =1/2*(C4B4) 4 =IF(E4<0,D4,B4) =IF(E4<0,C4,D4) =1/2*(B5+C5) =(D52)*(D54)*(D56)+1 =1/2*(C5B5) 5 =IF(E5<0,D5,B5) =IF(E5<0,C5,D5) =1/2*(B6+C6) =(D62)*(D64)*(D66)+1 =1/2*(C6B6) 6 =IF(E6<0,D6,B6) =IF(E6<0,C6,D6) =1/2*(B7+C7) =(D72)*(D74)*(D76)+1 =1/2*(C7B7) 7 =IF(E7<0,D7,B7) =IF(E7<0,C7,D7) =1/2*(B8+C8) =(D82)*(D84)*(D86)+1 =1/2*(C8B8) 8 =IF(E8<0,D8,B8) =IF(E8<0,C8,D8)
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Pure Mathematics 2: Solution of equation by Numerical Methods
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 midpoint 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 midpoint 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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Numerical Method (Maths Investigation)
EQUATION USED: = 0 has only 1 roots in [0,1] From Graphmatica: Results obtained from Microsoft Excel: Xvalue F(X) 1 1.5 0 0.5 Observe the change of yvalue from negative value to positive value in the table. That shows that the root of the equation where the yvalue is 0, is between x = 0 and x = 1. Now this is step 1, we're moving on to step 2. Xvalue 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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Repeated Differentiation
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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