Euler’s method and so on… Euler’s method of approximation of a function y = f(x) is approximations. constructed from repeated applications of linear It consists of generating a sequence of ordered pairs (x i, y i ). The sequence is generated in the following way.  (a, b) are the coordinates of the starting point  x 1 = ay 1 = b  x 2 = x 1 + hy 2  y 1 + f (x 1 ) × h where h is the step size  x 3 = x 2 + hy 3  y 2 + f (x 2 ) × h

Euler’s method Consider approximating the function f(x) = e x over the interval [2, 2.2].  Choose (2, e 2 ) as the coordinates of the starting point, i.e., (x 1, y 1 ) = (2, e 2 ). x y y = e x (2, e 2 )

Euler’s method  Choose a step size of 0.1, therefore h = 0.1 y x ( 2, e 2 ) ( 2.1, e 2.1 ) 0.1 f (x 1 ) × h = e 2 × 0.1 (  0.739) Gradient = e 2 The approximated value of the function when x = 2.1 is e e 2 = 1.1e 2 = e   x 2 = x 1 + h = = 2.1  f  (x) = e x, f  (x 1 ) = e 2.The straight line with gradient e 2 is drawn from the start point. (2.1, 8.128) The actual value of the function when x = 2.1 is e 2.1 (8.166…) y = e x

Euler’s method Gradient = e 2.1  The process can be repeated with a new start point at (2.1, 1.1e 2 ) y x f (x 2 ) × h = e 2.1 × 0.1 (  0.817) The approximated value of the function when x = 2.2 is 1.1e   x 3 = x 2 + h = = 2.2  f  (x 2 ) = e 2.1.The straight line with gradient e 2.1 is drawn from the new start point. (2.1, 1.1e 2 ) The actual value of the function when x = 2.2 is e 2.2 (9.025…) y = e x ( 2.2, e 2.2 ) (2.2, 8.945)

Euler’s method In summary: (x 1, y 1 ) = (2, e 2 ) Gradient = e 2 f (x 1 ) = e 2 (x 2, y 2 ) = (2.1, 8.128) y x y = e x ( 2.1, e 2.1 ) ( 2.2, e 2.2 ) f (x 2 ) = e 2.1 Gradient = e 2.1 (x 2, y 2 ) = (2.1, 8.128) (x 3, y 3 ) = (2.2, 8.945) (x 1, y 1 ) = ( 2, e 2 )