Numerical Methods

Numerical Methods

1 The root of the equation f(x) = 0, where is to be estimated using the iterative formula (a) Showing your values of x1, x2, x3,…, obtain the value, to 3 decimal places, of the root. (4) (b) By considering the change of sign of f(x) in a suitable interval, justify the accuracy of your answer to part (a). (2) Numerical Methods

1 The root of the equation f(x) = 0, where is to be estimated using the iterative formula Showing your values of x1, x2, x3,…, obtain the value, to 3 decimal places, of the root. (4) Numerical Methods

1 The root of the equation f(x) = 0, where is to be estimated using the iterative formula (b) By considering the change of sign of f(x) in a suitable interval, justify the accuracy of your answer to part (a) (2) The function is continuous and there is a change of sign, therefore x is a root Numerical Methods

2 f(x) = x3 + x2 4x  1. The equation f(x) = 0 has only one positive root,  (a) Show that f(x) = 0 can be rearranged as (2) The iterative formula xn + 1 = is used to find an approximation to . • Taking x1 = 1, find, to 2 decimal places, the values of • x2, x3 and x4. (3) Numerical Methods

2 f(x) = x3 + x2 4x  1. The equation f(x) = 0 has only one positive root,  (a) Show that f(x) = 0 can be rearranged as (2) Numerical Methods

2 f(x) = x3 + x2 4x  1. The equation f(x) = 0 has only one positive root,  • Taking x1 = 1, find, to 2 decimal places, the values of • x2, x3 and x4. (3) Numerical Methods

2 f(x) = x3 + x2 4x  1. The equation f(x) = 0 has only one positive root,  (c) By choosing values of x in a suitable interval, prove that  = 1.70, correct to 2 decimal places. (3) • Write down a value of x1 for which the iteration formula • xn + 1 = • does not produce a valid value for x2. • Justify your answer. (2) Numerical Methods 2 4.

2 f(x) = x3 + x2 4x  1. The equation f(x) = 0 has only one positive root,  (c) By choosing values of x in a suitable interval, prove that  = 1.70, correct to 2 decimal places. (3) f(1.695) = -0.037… f(1.705) = 0.0435… The function is continuous and there is a change of sign, therefore  is a root Numerical Methods

2 f(x) = x3 + x2 4x  1. The equation f(x) = 0 has only one positive root,  • Write down a value of x1 for which the iteration formula • xn + 1 = • does not produce a valid value for x2. • Justify your answer. (2) x = -1; Division by zero is not possible -1 < x < -¼; Cannot find the square root of a negative number Numerical Methods

3 (a) Sketch, on the same set of axes, the graphs of y = 2 – ex and y = x. (3) [It is not necessary to find the coordinates of any points of intersection with the axes.] Given that f(x) = ex + x – 2, x 0, (b) explain how your graphs show that the equation f(x) = 0 has only one solution, (1) Numerical Methods

3 (a) Sketch, on the same set of axes, the graphs of y = 2 – ex and y = x. (3) [It is not necessary to find the coordinates of any points of intersection with the axes.] Numerical Methods

3 Given that f(x) = ex + x – 2, x 0, (b) explain how your graphs show that the equation f(x) = 0 has only one solution, (1) Where curves meet is solution to f(x) = 0; hence there is only one intersection Numerical Methods

3 (c) show that the solution of f(x) = 0 lies between x = 3 and x = 4. (2) The iterative formula xn + 1 = (2 – )2 is used to solve the equation f(x) = 0. (d) Taking x0 = 4, write down the values of x1, x2, x3 and x4, and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. (4) Numerical Methods

3 (c) show that the solution of f(x) = 0 lies between x = 3 and x = 4. (2) f(x) = ex + x – 2 f(3) = -0.218… f(4) = 0.018… This is a continuous function and there is a change of sign over this interval, therefore there is a root within this interval. Numerical Methods

The iterative formula xn + 1 = (2 – )2 is used to solve the equation f(x) = 0. (d) Taking x0 = 4, write down the values of x1, x2, x3 and x4, and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. (4) 3 x0 = 4 x1=(2 – e–4)2 = 3.92707… x2 = 3.92158… x3 = 3.92115… x4 = 3.92111(9)… Approx. solution = 3.921 (3 dp) Numerical Methods

4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). (a) Sketch the graph of y = ln 3x, showing the exact value of p (2) The normal to the curve at the point Q, with x-coordinate q, passes through the origin. Show that x = q is a solution of the equation x2+ ln 3x = 0 (4) Numerical Methods 5 8.

4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). (a) Sketch the graph of y = ln 3x, showing the exact value of p (2) ln 1 = 0 ln3x = ln1 3x = 1 x = 1/3 Numerical Methods 5 8.

4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). The normal to the curve at the point Q, with x-coordinate q, passes through the origin. Show that x = q is a solution of the equation x2+ ln 3x = 0 (4) y = ln3x, y’ = 1/x At Q gradient is 1/q Gradient of normal is -q Equation of OQ is y – 0 = -q(x- 0) y = -qx ln 3x = -x2 Q  x2+ ln 3x = 0 Numerical Methods 5 8.

4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). • Show that the equation in part (b) can be rearranged in the form x = (2) • (d) Use the iteration formula xn + 1 = , with x0 = , to find x1, x2, x3 and x4. Hence write down, to 3 decimal places, an approximation for q(3) Numerical Methods 5 8.

4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). • Show that the equation in part (b) can be rearranged in the form x = (2) x2+ ln 3x = 0 ln 3x = - x2 3x = x =1/3e-x2 Numerical Methods 5 8.

4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). (d) Use the iteration formula xn + 1 = , with x0 = , to find x1, x2, x3 and x4. Hence write down, to 3 decimal places, an approximation for q(3) x1 = 0.298280; x2= 0.304957, x3= 0.303731, x4= 0.303958 Root = 0.304 (3 decimal places) Numerical Methods 5 8.

Numerical Methods