Solutions To Numerical Analysis Problems (3)

 1)  Find the x coordinate of the point where the curve  x ³  - x  crosses the horizontal line y = 1.

Solution:

The given curve crosses the line when:

 x ³  - x  =   1

Or:   x ³  - x  - 1   =  0

We then ask: When does  x ³  - x  - 1   =  0

The graph (below) shows a single root located between x = 1 and x = 2:

Try: x = 1.3 =  x _n 

For Newton's Method:

x _n+1  =   x_n  -  f(x_n) / f '(x_n) 

= 1.3 - {(1.3) ³    - (1.3) - 1)}/ 3 (1.3))²  -  1 =  1.325

Which turns out to be very close to the actual root value.

2) Using Newton's method find at least one root approximation for the equation:

x ³   =   2    Rewrite as: f(x) =   x ³  -   2  = 0  

We use a graph for f(x)  =   x ³  -   2, e.g.:

And note a likely root occurs at about: 1.2, so try: x = 1.2 =  x _n 

For Newton's Method:

x _n+1  =   x_n  -  f(x_n) / f '(x_n) 

Based on:

f(x) = x ³  -   2  and f'(x) = 3x ²

x _n+1  =  1.2 - (x ³  -   2 ) / 3x ²

= (1.2) - {(1.2) ³  -  2)}/ 3 (1.2))²  = 1.263

Since:  f '(x_n)  =  3x ²

We see the approximated value is close enough. Check:

x ³  -   2  = 0    vs.

(1.263) ³   - 2  = 0.015