3.
$$\mathrm { f } ( x ) = \frac { x ^ { 2 } } { 4 } + \ln ( 2 x ) , \quad x > 0$$
- Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as
$$x = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x ^ { 2 } }$$
The equation \(\mathrm { f } ( x ) = 0\) has a root near 0.5
- Starting with \(x _ { 1 } = 0.5\) use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x _ { n } ^ { 2 } }$$
to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
- Using a suitable interval, show that 0.473 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.