AS and A Level: Core & Pure Mathematics
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Am going to use numerical methods to solve equations that can't be solved algebraically
I expand this again twice more to find the 3rd and 4th decimal places which are (0.716 & 0.717) and (0.7169 and 0.717) respectively. 0.7 0.148 0.71 0.06057 0.72 0.026112 0.73 0.111998 0.74 0.197056 0.716 0.00847 0.75 0.28125 0.7161 0.0076 0.76 0.364544 0.7162 0.00673 0.77 0.446902 0.7163 0.00587 0.78 0.528288 0.7164 0.005 0.79 0.608666 0.7165 0.00414 0.8 0.688 0.7166 0.00327 0.7167 0.00241 0.7168 0.00154 0.71 0.06057 0.7169 0.00068 0.711 0.05186 0.717 0.00019 0.712 0.04317 0.713 0.03448 0.714 0.0258 0.71695 0.00024 0.715 0.01713 0.71705 0.000623 0.716 0.00847 0.717 0.00019 0.718 0.008839 0.719 0.017479 0.72 0.026112 0.71695 0.00024 0.71705 0.000623 Thus proves that the root lies no greater nor less than 0.00005 away from 0.717.
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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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In this coursework I am going to investigate the relationship between the orbital period and the distance of the planet from the sun. Assume that this relationship is a power law of the form: T = KR^n
After separating the logs we must use the log rule number 3, which is Log x+ k log x. Keep in mind that the transposition should end up to the linear function Y=c + mx. * Log k as C, which is the intercept of the line * N as the gradient of the line and an exponent in the formula * Log R as the as the base of the exponent n Log T = log K + n log R is the same as Y = c + mx I've calculated the log of R
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Mathematical Investigation
by varying values of "a" only changes the standard sine function y= a�sin(x)'s amplitude according to a. Varying values of "a" does not change the function's amplitude or X and Y intersects. When the value of "a" increases the function is stretched vertically and so the amplitude increases as well. Vise versa, when the value of "a" decreases, the function is vertically compressed and so the amplitude decreases as well. (b) Based upon the properties observed from above figures, varying values of "a" appear only change the height (amplitude) of wave pattern and nothing else. Different values of "a" does not effect the period as the period in all graphs remain the same.
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Henna Night
At a Henna function it is traditional for the bride to wear the colour yellow or golden. The drole of the drum was getting louder and louder, with the beat gradually getting faster. There were two lines of girls waiting for Yasmin to pass by. They were holding little baskets with party poppers and flowers inside. Included in this line Nadia and her cousins. People looked left and right, up and down, to see where the beat of the drum was coming from. The drummer was at the front, slowly walking and behind was Yasmin. A yellow shimmering sheet with golden threads at each corner of the sheet was carried over Yasmin's head by her two brothers, Wasim and Aseem.
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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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Solving the equation of 0 = 3x^5  3x + 1 using different methods
0.333 0.013284074 0.334 0.010469631 0.335 0.007657423 0.336 0.004847471 0.337 0.002039795 0.338 0.000765585 0.339 0.003568648 0.34 0.006369373 x y 0.3377 7.57868E05 0.33771 4.77377E05 0.33772 1.96889E05 0.33773 8.35975E06 0.33774 3.64081E05 0.33775 6.44563E05 0.33776 9.25042E05 0.33777 0.000120552 0.33778 0.000148599 0.33779 0.000176647 0.3378 0.000204694 x y 0.337 0.002039795 0.3371 0.001759153 0.3372 0.001478535 0.3373 0.001197939 0.3374 0.000917366 0.3375 0.000636817 0.3376 0.00035629 0.3377 7.57868E05 0.3378 0.000204694 0.3379 0.000485151 0.338 0.000765585 Using the Decimal Search/Change of Sign method, it took 32 calculations to reach a conclusive answer to 5 decimal places of 0.337735 �0.000005.
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Graph Original of Equation y=3x^53x+1
When it reaches the second curve, another line is drawn horizontally. This carries on until a value of the function is calculated to the required accuracy.
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GCSE Math Coursework: Triminoes
3. I will continue to do this until I feel I have used enough numbers. 4. On each set of data I will record the largest number used, the sum of all the cards added together, and the number of cards used. 5. I will then make a results table. 6. I will collect the data from each group and place it together then I will find out whether the equation is a linear, quadratic, cubic or quartic. 7. By replacing the correct numbers with the correct letters I will find out the formula. 8. I will then write down the formula for each set of sequence.
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Borders Coursework
These values will then be substituted in the equation with 'n' to check that the formulas are working. These are the 'values of n'. All formulas that I will be calculating will be tested to make sure that they are correct. Part 1: 2D Shapes: 1st term: 2nd term: 3rd term: 4th term: 5th term: 6th term: 7th term: 8th term: 9th term: Part 1: 2D Shapes: Table to show the amount of squares in total and the amount of extra squares used for the 2D drawing: Value of 'n' 1 2 3 4 5 6 7 8 9 Extras
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Sequences and series investigation
2(3 squares) 1(5 squares) The Patterns I Have Noticied in Carrying Out the Previous Method I have now carried out ny first investigation into the pattern and have seen a number of different patterns. Firstly I can see that the number of squares in each pattern is an odd number. Secondly I can see that the number of squares in the pattern can be found out by taking the odd numbers from 1 onwards and adding them up (according to the sequence).
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Methods of Advanced Mathematics (C3) Coursework.
6 183 5 97 4 41 3 9 2 5 1 7 0 3 1 1 To home in on a more accurate answer I need to investigate the decimal places between 3 and 2 x f(x) 3 9 2.9 6.889 2.8 4.952 2.7 3.183 2.6 1.576 2.5 0.125 2.4 1.176 2.3 2.333 2.2 3.352 2.1 4.239 2 5 This method is repeated until a sufficient number of decimal places are achieved. In this investigation I think 5 decimal places will be enough 2.5 0.125 2.49 0.011751 2.48 0.147008 2.47 0.280777 2.46 0.413064 2.45 0.543875 2.44 0.673216 2.43 0.801093 2.42
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Der Handschuh.
6 183 5 97 4 41 3 9 2 5 1 7 0 3 1 1 To home in on a more accurate answer I need to investigate the decimal places between 3 and 2 x f(x) 3 9 2.9 6.889 2.8 4.952 2.7 3.183 2.6 1.576 2.5 0.125 2.4 1.176 2.3 2.333 2.2 3.352 2.1 4.239 2 5 This method is repeated until a sufficient number of decimal places are achieved. In this investigation I think 5 decimal places will be enough 2.5 0.125 2.49 0.011751 2.48 0.147008 2.47 0.280777 2.46 0.413064 2.45 0.543875 2.44 0.673216 2.43 0.801093 2.42
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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(62x)^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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Estimate a consumption function for the UK economy explaining the statistical techniques you have used.
It is thus very important to forecasters for them to be able to predict consumption correctly. Even a small percentage of error can result in a very large absolute error. For example, forecasters make an error of 1 per cent in predicting consumption, which seems to be insignificant. This will account for an error of 0.7 per cent of GDP. Therefore its accurate prediction is essential to the management of the economy. If we can model the aggregate consumption function, then we can go along predicting future consumption level and use fiscal and monetary policy to manage the economy efficiently.
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Investigate the relationships between the lengths of the 3 sides of the right angled triangles and the perimeters and areas of these triangles.
4 , 12 , 24 , 40 8 , 12 , 16 4 , 4 5 , 13 , 25 , 41 8 , 12 , 16 4 , 4 Length of shortest side: Term no 1 2 3 4 5 Sequence 3 5 7 9 11 Sequence 2n 1 4 6 8 10 Sequence 1 1 1 1 1 2n + 1 Length of the middle side: F (n) = an� + bn +c F= (1) = a x 1� + b x 1 + c = a + b + c = 4  eqn1 F= (2)
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Numerical Solutions of Equations
To find the smallest root, I will take x = 4 as my starting value, and see the number of iterations required to find the root. x Y=x^52.7x+1.8 1 3.5 1.1 3.15949 1.2 2.55168 1.3 1.59707 1.4 0.20176 1.5 1.74375 1.6 4.36576 1.7 7.80857 1.8 12.23568 1.9 17.83099 2 24.8 x Y=x^52.7x+1.8 1.4 0.20176 1.41 0.03391633 1.42 0.139533923 1.43 0.318710894 1.44 0.503736422 1.45 0.694734062 1.46 0.891829098 1.47 1.095148551 1.48 1.304821197 1.49 1.520977575 1.5 1.74375 x Y=x^52.7x+1.8 1.411 0.01682557 1.4111 0.015113406 1.4112 0.013400679 1.4113 0.011687391 1.4114 0.009973541 1.4115 0.008259128 1.4116 0.006544153 1.4117 0.004828615 1.4118 0.003112515 1.4119 0.001395852 1.412 0.000321374 x Y=x^52.7x+1.8
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Using Decimal search
 Graph 3  The graph shows me that 10x^32.5x+0.2=0 has three roots, lying in the intervals (1,0), (0,1). X 1 0 1 Y 7.3 0.2 7.7 I can stop here as there is no change of sign in interval (0,1). Even though no root is indicated, the graph shows 2 roots which this method fails to retrieve. Using the Newton Raphson Iteration Method For a graph that works using Newton Raphson method. Xn+1 = Xn  f (Xn) f'(Xn) Where f (X1) = x^3+3x^2+1.4x1 f '(X) = dy = 3x^2+6x+1.4 dx The following graph represents y=f(x)  Graph 4  The graph shows me that x^3+3x^2+1.4x1 =0 has 3 roots, lying in the intervals (3,2)
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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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Change of Sign Method
x 1.78 1.779 1.778 1.777 1.776 1.775 1.774 1.773 1.772 1.771 1.77 y 0.409 0.3659 0.3228 0.2799 0.2371 0.1944 0.1519 0.1094 0.067 0.0248 0.0173 Therefore there is a root between x=1.771 and x=1.770 x 1.771 1.7709 1.7708 1.7707 1.7706 1.7705 1.7704 y 0.0248 0.0206 0.0164 0.0121 0.0079 0.0037 0.0005 This shows that the root between x=2 and x=1 is between x=1.7705 and x=1.7704. The accuracy of this method to find roots is stated below. 1.7705 < x < 1.7704 x = 1.770 (3 d.p.)
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Experimentally calculating the wavelength of an HeNe laser by means of diffraction gratings
In the case of constructive interference: dsin? = m?, where d is the distance between the slits, ? is the angle at which the light is diffracted, m is the fringe order being considered, and ? is the wavelength of the light. In the case of this experiment, a monochromatic HeNe light (basically a HeNe laser) is being pointed through "diffraction gratings," which are defined as "large numbers of equally spaced parallel slits."2 To find the wavelength of the light experimentally, one can use the same equation for the interference pattern from double slits. One is given the distances between the slits (which would be the same as 1 divided by the number of slits).
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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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Analyse the use of three methods which are called the: change of sign, NewtonRaphson 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+4x46x 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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Investigate the three different numerical methods used to solve equations.
Y=x35x+1 can be solved using decimal search. The table shows that the routes can be found. X Y X Y X Y 0 1 2 1 2 3 0.1 0.501 2.1 0.239 2.1 2.239 0.2 0.008 2.2 0.648 2.2 1.352 0.3 0.473 2.11 0.15607 2.3 0.333 0.21 0.04074 2.12 0.07187 2.4 0.824 0.201 0.003121 2.13 0.013597 2.31 0.223609 0.202 0.00176 2.121 0.06338 2.32 0.112832 0.2011 0.002633 2.122 0.05488 2.33 0.000663 0.2012 0.002145 2.123 0.04637 2.34 0.1129 0.2013 0.001657 2.124 0.03784 2.331 0.01063 0.2014 0.001169 2.125 0.0293 2.3301 0.00047 0.2015 0.000681 2.126 0.02074 2.33001 0.00055 0.2016 0.000194 2.127 0.01218 Root= 2.33005
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Numerical Methods used to solve those equations which cannot be solved analytically.
Therefore the root of the equation is 1.685 � 0.05 Third Interval x f(x) 1.68 0.01101 1.681 0.009628 1.682 0.008243 1.683 0.006854 1.684 0.005459 1.685 0.004059 1.686 0.002654 1.687 0.001243 1.688 0.0001725 1.689 0.001593 1.69 0.003019 Therefore, we can say that the one root of the equation; y=0.6938x3  0.9157x2  1.421x + 1.671 is 1.6885 � 0.0005 However, decimal search cannot be used in all circumstances. This can be demonstrated using the equation: y=0.6918x3  0.2159x2  3.019x + 2.77 GRAPH If we zoom in, the roots of the graph can be seen more clearly: TABLE OF VALUES x f(x)
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