Sketch the graphs of y=lnxy = \ln xy=lnx and y=1xy = \dfrac{1}{x}y=x1 for x>0 x > 0\,x>0 on the same axes.
Explain why the equation lnx=1x\ln x = \dfrac{1}{x}lnx=x1 has exactly one solution.
By considering a suitable change of sign, show that the solution lies between 1.7 and 1.8.
Use the iterative formula
xn+1=e1xn\displaystyle x_{n+1} = e^{\frac{1}{x_n}}xn+1=exn1
with x1=1.8x_1 = 1.8x1=1.8 to find x2x_2x2, x3 x_3\,x3 and x4x_4x4, giving your answers to four decimal places.
137 exam-style questions on OCR (MEI) A Level Maths 1.10 Numerical Methods (A-level only), covering 1.10.1 Locate roots by change of sign (A-level only), 1.10.2 When change of sign methods fail (A-level only), 1.10.3 Fixed point iteration (A-level only), 1.10.4 Newton-Raphson method (A-level only), 1.10.5 Convergence of iterations (A-level only), 1.10.6 Trapezium rule (A-level only), 1.10.7 Upper and lower bounds using rectangles (A-level only), and 1.10.8 Numerical methods to solve problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.