The equilibrium yield YYY of a specific chemical synthesis is modeled by the equation Y3−6Y+3=0Y^3 - 6Y + 3 = 0Y3−6Y+3=0. This equation has a single positive solution, Y=αY = \alphaY=α, in the interval [2.1,2.2][2.1, 2.2][2.1,2.2].
By considering a suitable change of sign, show that α\alphaα lies between 2.1 and 2.2.
Show that the equation Y3−6Y+3=0Y^3 - 6Y + 3 = 0Y3−6Y+3=0 can be rearranged into the form
Y=6−3Y Y = \sqrt{6 - \frac{3}{Y}} Y=6−Y3Use the iterative formula
Yn+1=6−3Yn Y_{n+1} = \sqrt{6 - \frac{3}{Y_n}} Yn+1=6−Yn3with Y1=2.1Y_1 = 2.1Y1=2.1, to determine the values of Y2,Y3Y_2, Y_3Y2,Y3 and Y4Y_4Y4, giving your answers to four decimal places.
Hence, deduce an interval of width 0.001 in which α\alphaα lies.
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.