How to find a cube root geometrically?

Apply Newton's method to find a root of f(x)=x^3-10 in the interval (2,3). If your initial guess is x0=2, then approximately how far is x2 from the cube root of 10 (the actual root of the equation)?

• Answer:

  About this: X[2] - X[9] 2.15450825808003 - 2.15443469003188 = 0.00007356804815 (More exactly, cbrt(10) = 2.1544346900318837217592935665195) "X[1] = 2.1665833611042 X[2] = 2.15450825808003 X[3] = 2.15443472667828 X[4] = 2.15443469004889 X[5] = 2.15443469003189 X[6] = 2.15443469003188 X[7] = 2.15443469003188 X[8] = 2.15443469003188 X[9] = 2.15443469003188" Source and further information: http://www.math.sc.edu/cgi-bin/sumcgi/Newton.pl "Given a function ƒ(x) and its derivative ƒ '(x), we begin with a first guess x0 . A better approximation x1 is x1 = x0 - f(x0)/f'(x0) Newton's method is an extremely powerful technique -- in general the convergence is quadratic: the error is essentially squared at each step (that is, the number of accurate digits doubles in each step). However, there are some difficulties with the method. 1. Newton's method requires that the derivative be calculated directly. In most practical problems, the function in question may be given by a long and complicated formula, and hence an analytical expression for the derivative may not be easily obtainable. In these situations, it may be appropriate to approximate the derivative by using the slope of a line through two points on the function. In this case, the Secant method results. This has slightly slower convergence than Newton's method but does not require the existence of derivatives. 2. If the initial value is too far from the true zero, Newton's method may fail to converge. For this reason, Newton's method is often referred to as a local technique. Most practical implementations of Newton's method put an upper limit on the number of iterations and perhaps on the size of the iterates. 3. If the derivative of the function is not continuous the method may fail to converge. 4. It is clear from the formula for Newton's method that it will fail in cases where the derivative is zero. Similarly, when the derivative is close to zero, the tangent line is nearly horizontal and hence may "shoot" wildly past the desired root. 5. If the root being sought has multiplicity greater than one, the convergence rate is merely linear (errors reduced by a constant factor at each step) unless special steps are taken. When there are two or more roots that are close together then it may take many iterations before the iterates get close enough to one of them for the quadratic convergence is apparent." Source and further information: http://en.wikipedia.org/wiki/Newton's_method In the given example, f(x) = x^3-10 f'(x)= 2*x^2 x1 = x0 - ([x0]^3 - 10)/ (2*[x0]^2)