The method I am going to use to solve x−3x-1=0 is the Change Of Sign Method involving the Decimal Search method

Mathematics C3 Coursework Numerical solutions of equations Decimal Search Method --------------------------------------------p.1 Rearrangement Method ------------------------------------------- p.5 Newton-Raphson Method ----------------------------------------- p.9 Comparison ----------------------------------------------------------- p.12 Solving x³-3x-1=0 using the "Change Of Sign" Method: The method I am going to use to solve x³-3x-1=0 is the Change Of Sign Method involving the Decimal Search method, which is that you are looking for the roots of the equations f(x)=0. This means that I want the value of x for which the graph of y=f(x) crosses the x axis. As the curve crosses the x axis, f(x)changes sign, so provided that f(x) is continuous function, once I have located an interval in which f(x) changes sign, I know that that interval must contain a root. Now, I have drawn the graph of x³-3x-1=0 by using the Autograph software, and the graph is shown below: The point that the arrow pointing is the root I need to find. From my graph above, I can see that the root of the equation is between x=0 and x=-1. The table of x values and f(x) values is shown below. I can work out the f(x) values by substituting the x-values into the equation. x f(x) -0.1 -1.299 -0.2 -0.408 -0.3 -0.127

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  • Level: AS and A Level
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Investigation of circumference ratio - finding the value of pi.

Stoney Li 11g HL math Explain why the marked angle is 30 In this diagram, the C point is the center of the circle and hexagon. The triangle ACD is one sixth of the hexagon. So the angle ACD is one sixth of central angle. The central angle is 360 degree, which mean the angle ACD is 60 degree. The segment BC connects the midpoint of segment AB to the point C. So segment AB equal segment BD. Because the segment AC and segment CD are radius of circle, the triangle ACD is isosceles triangle. That mean segment BC divides the triangle ACD to two triangles and this two triangles are equal. Which mean angle ACD had divided to angle ACB and angle BCD. So 60-degree divide by 2, the angle ACB equal angle BCD equal 30 degree. Use trigonometric ratio(s) to find the area of the hexagon. At first we find the area of triangle ACD. 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. 2-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

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  • Level: AS and A Level
  • Subject: Maths
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Pure Mathematics 2: Solution of equation by Numerical Methods

Pure Mathematics 2: Solution of equation by Numerical Methods Introduction: In this coursework, I am going to solve equations by using the Numerical Methods. Numerical methods are used to solve equation that cannot be solved algebraically e.g. quadratic equations ax²+bx+c=0 can be solved using this formula: x= -b± V b² - 4ac 2a Therefore numerical methods would not be used for quadratic equations. I will be working with cubic equation because there is no formula to solve it. There are three methods, which I will be using: * Change of sign method * Newton-Raphson method * Rearranging f(x) = 0 in the form x = g(x) Change of sign method: This method is concerned with when a function crosses the x-axis, and by definition changes sign (+ and -). If we are looking the root of equation f(x) = 0. The point at which the curve crosses x-axis is the root. Once an interval where f(x) changes sign then the root must be in the interval. f(a) > 0 f(b) < 0 Therefore root must be between [a,b] f(a) < 0 f(b) > 0 Root is between the interval [a,b] To find the interval of each root for the equation, I'll be doing a decimal search first. Lets take the equation y = x³ - 12x + 5 x -4 -3 -2 -1 0 2 3 4 f(x) -11 4 21 6 5 -6 -11 -4 21 There are 3 roots in this equation and they are in these intervals: [-4, -3] [ 0 , 1 ] [ 3 , 4 ] Now, I am going to use

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  • Level: AS and A Level
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'Change of Sign Method'

MATHEMATICS COURSEWORK Numerical Methods can be used to solve those equations which cannot be solved analytically. One such method is the 'Change of Sign Method', focusing on Decimal Search. Decimal search is based on the principal that f(x) changes sign as a curve passes through the x axis at a root. Above the x axis, f(x) is positive and below the x axis f(x) is negative. Using this principal, values of f(x) are initially found and consecutive values are found in increments of 0.1 until a change in sign is found. If further accuracy is required, consecutives values could then be found in increments of 0.01 and 0.001, until a change of accuracy is found, therefore giving one root of the equation. EVIDENCE Equation: y=0.6938x3 - 0.9157x2 - 1.421x + 1.671 GRAPH: TABLE OF RESULTS First Interval x f(x) 0.0281 .1 -0.07665 .2 -0.1539 .3 -0.1996 .4 -0.2094 .5 -0.1792 .6 -0.105 .7 0.01757 .8 0.1926 .9 0.4242 The table shows that the sign changes within the interval 1.6 to 1.7. This tell us that the roots lie somewhere within this interval; Therefore the root is 1.65 ± 0.05 Second Interval X f(x) .6 -0.105 .61 -0.09497 .62 -0.08447 .63 -0.07348 .64 -0.06199 .65 -0.05001 .66 -0.03752 .67 -0.02452 .68 -0.01101 .69 0.003019 The table of values shows that the root lies in the interval [1.68, 1.69]. Therefore the root of the

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  • Level: AS and A Level
  • Subject: Maths
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Change of Sign Method.

Change of Sign Method If we let f(x)?3x³-0.5x²-0.5x-1, in order to solve this equation and determine its roots, it is necessary for it to be written in the form 3x³-0.5x²-0.5x-1=0. The root of the equation f(x)=0 is indicated where y=f(x) crosses the x-axis. Roots of the equation 3x³-0.5x²-0.5x-1=0, will be found to a three decimal place accuracy. Having illustrated the equation graphically using Autograph, it is evident that the equation has only root that lies between the interval [0,1]. Before proceeding, it is necessary to check that there is a sign change in the above interval: f(0) = 0-0-0-1= -1 f(1) = 3-0.5-0.5-1 =1 The method that will be used for the numerical solution of the equation is the decimal search method. I will first take the increments in x of size 0.1 within the interval [0,1], calculating the value of the function 3x³-0.5x²-0.5x-1 for each one, until a change of sign is found. x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 f(x) -1 -1.052 -1.096 -1.114 -1.088 -1 -0.832 -0.566 -0.184 0.332 The above table illustrates that there is a sign change. It can therefore be understood that the root lies in the interval [0.8,0.9]. Having narrowed down the interval, I will continue with the decimal search, but now using increments of 0.01 within the interval [0.8,0.9]. x 0.80 0.81 0.82 0.83 0.84 f(x) -0.184 -0.138727

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decimal search

Introduction In maths equations can be solved using various methods. A very common and efficient method in solving equations is algebraically. But not all equations can be solved algebraically; these equations must be solved using numeric methods. I will study three specific numeric methods on different equations. ~ Change of sign, decimal search process. ~ Newton-Raphson method. ~ Re-arrangement method. I will be testing the numeric methods with separate equations which cannot be solved algebraically. I will also apply all of the methods to one of the equations and check if all the methods give me the same value for the root I want to find. Change of Sign, Decimal Search To find the root of the equation f(x) = 0 means finding values of x for the graph y = f(x). The change of sign method works on the bases that the y = f(x) graph changes signs when it crosses the x-axis. e.g. y = f(x) The sketch above shows that there is a root between the interval [b , c] and the curve of y = f(x) crosses the x-axis and changes its sign from negative ( - ) to positive ( + ), and at the interval [a , b] f(x) curve crosses the x-axis changing its sign from positive ( + ) to negative ( - ). An initial interval of where a root lies can be obtained from a sketch. By taking the values of the initial interval we can increase the value of x by increments of 0.1 within the

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C3 COURSEWORK - comparing methods of solving functions

Change of signs First of all, I would like to use change of signs method to find the roots of the function y=x³+3x²–3. It is presented by the graph of y=f(x) y= x³+3x²–3 By using the change of signs method, we can find out that there are 3 changes of signs in the range [-3, -2], [-2, -1] and [0, 1] x³+3x²–3 -3 -3 Change of sign -2 1 -1 -1 Change of sign 0 -3 Change of sign 1 1 2 17 The first 2 calculations of x³+3x²–3: The change of sign here tells us that there is a root in the interval [0, 1], where roots are not trivial. (f (x) not equal to zero) Roots to f(x) =0 y= x³+3x²–3 x x³+3x²–3 0 -3 0.1 -2.969 0.2 -2.872 0.3 -2.703 0.4 -2.456 0.5 -2.125 0.6 -1.704 0.7 -1.187 0.8 -0.568 Change of sign 0.9 0.159 1 1 The change of sign here tells us that there is a root in the interval [0.8, 0.9]. x y=x³+3x²–3 0.8 -0.568 0.81 -0.50026 0.82 -0.43143 0.83 -0.36151 0.84 -0.2905 0.85 -0.21838 0.86 -0.14514 0.87 -0.0708 Change of sign 0.88 0.004672 0.89 0.081269 0.9 0.159 The change of sign here tells us that there is a root in the interval [0.87, 0.88]. x y=x³+3x²–3 0.87 -0.0708 0.871 -0.0633 0.872 -0.05579 0.873 -0.04827 0.874 -0.04074 0.875 -0.0332 0.876 -0.02565 0.877 -0.01809 0.878 -0.01051 0.879 -0.00293 Change of sign 0.88 0.004672 The change of

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  • Level: AS and A Level
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Newton Raphson Method for Solving 6x3+7x2-9x-7=0

Page of Newton-Raphson method The equation I am going to Solve is 6x3+7x2-9x-7=0 Newton-Raphson is a fixed point iteration method. The first approximation has to be near the root so the method will work. A tangent has to be drawn at the first point y=f(x) and when it touches the x-axis this gives another root. The tangent cuts the second point on the curve. This process is then repeated. The Newton Raphson method uses an algebraically formula to calculate the value of roots. . The formula is first worked out by the gradient of the tangent there are steps of algebra to get to the final equation. Autograph will do these calculations automatically. f `(xn) = (x1-x2)f `(x1) = f(x1) x1-x2 = x2 =x1 - f(x) is negative The root lies between 1.1155 and 1.11555 .11545 1.1155 1.11555 f(1.11545) = -0.002199645 f(1.1155) = 0.00020309 f(1.11555)= 0.000701867. X = 1.1155 Error Bounds = ±0.00005 Solution bounds = (1.11545≤ x ≤1.11555). n xn f(xn) f `(xn) xn+1 0 x0 = 1 f(x0)= -3 f '(x0)= 23 x1 = 1.1304 x1 = 1.1304 f(x1) = 0.4376 f '(x1) = 29.826 x2 = 1.1157 2 x2 = 1.1157 f(x2) = -0.000748 f '(x2) = 29.015 x3 = 1.1155 Solving 6x3+7x2-9x-7=0 for

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  • Level: AS and A Level
  • Subject: Maths
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Maths Assignment - trigonometry, trend line, probability and calculus questions.

. Question: Trigonometry can be used to solve problems instead of constructing a scale diagram. A man leaves a point walking at 6kmh on a bearing of 80o. A cyclist leaves the same point at the same time on a bearing of 120o travelling at a speed of 17kmh. Calculate their distance apart after 4 hours using trigonometry. (6) Answer: Figure 01 As shown in the figure 01, let’s take ‘A’ as the starting point. After 4hrs the man is at point M and cyclist at point C Distance travelled by the man after 4hrs Distance travelled by the cyclist after 4hrs Now let’s consider the triangle Using the information in the question, the application of the Cosine rule would seem to be the best choice for this problem as we are given more angles than distances. After 4hrs distance between the cyclist and man 2. Question: If experimental data doesn’t produce a straight trend line it doesn’t mean there isn’t a relationship. The use of software packages can allow quick and accurate testing of data to different trend lines so that a suitable relationship can be found if it exists. Atmospheric pressure P is measured at varying altitudes h as shown below. The data is thought to be of the exponential form p = aekh Altitude 500 1500 3000 5000 8000 Metres Pressure 73 68 62 54 43 cm a) Using Excel verify this is true.

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  • Level: AS and A Level
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The open box problem

The open box problem An open box is to be made from a sheet of card. Identical squares are to be cut off at the corners so that the card can be folded into the open box. The diagram below shows the sheet of card and the four corners, which are to be cut off. There are two objectives I shall be investigating: . To determine the size of the four square cuts that will make the volume of the box as large as possible with any given square sheet of card. 2. To determine the size of the four square cuts that will make the volume of the box as large as possible with any given rectangular piece of card. First I shall investigate objective 1 as I think it will be easier to do. Objective 1: The square sheet of card. There are 2 ways to solve this problem: I can use algebra, or I can use trial and improvement. I think I will start by using trial and improvement; I will construct a series of tables and graphs and see if I can find any patterns. If so, then I will be able to come up with a hypothesis, which I can then test to see if I can solve this first objective. Then I will attempt to solve the same problem using algebra. Method 1: Trial and improvement To do this I need to make up some dimensions and then apply them to the square. I will choose the dimensions 6x6 to start with. I will call the length of the square cuttings x. The second drawing shows the open box with the

  • Word count: 4307
  • Level: AS and A Level
  • Subject: Maths
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