0

Let $g(x) = x^2 \ln (x)$ and consider the equation $g(x) = a$ with $a = \frac{-3}{2e^3}$. The solution of this equation is given by $x = e^{-3/2}$. Note that $g''(e^{-3/2}) = 0$ and on a neighbourhood of $e^{-3/2}$ we have $f(x)f''(x) < 0$ with $f(x) := g(x) - a$. Show that NR cannot converge monotonically.

So $f(x)f''(x) < 0$ implies that either $f(x) > 0$ and $f''(x) < 0$ or $f(x) < 0$ and $f''(x) > 0$. If $f''(x) < 0$ then $f'(x)$ is decreasing and if $f''(x) > 0$ then $f'(x)$ is increasing. All this leads me to believe that NR will 'overshoot' the solution $x = e^{-3/2}$ regardless of where we start with our initial guess $x_0$ but I have no clue how to prove this. Any help is appreciated.

user119470
  • 59
  • 7

2 Answers2

0

This is not exact solution to your question but it will solve your problem. You have made it a little complicated.

Hint:

Here you are trying to solve the equation $f(x) = g(x) - a = 0$ where $g(x) = x^2 \ln(x)$.

Calculate $f'(x)$ and $f''(x)$.

Condition for convergence for Newton-Raphson method is $|f(x)f''(x)| \le |f'(x)|^2$

Put the values in the equation for any $a \neq -\frac{3}{2e^3}$. It will give convergence.

For $a = -\frac{3}{2e^3}$, see $g''(x) = 0$ gives $f''(x) = 0$ and the left hand side of the equation is $0$.

That gives the iteration of N-R method always converge.

Supriyo
  • 6,119
  • 7
  • 32
  • 60
0

If I properly understood, the question you ask is related to the overshoot of the solution depending on the starting point.

Graph the function $|[f(x) f''(x)] - [f'(x)]^2|$ as suggested by Hopeless Fool; you will notice that, in particular, it is negative for $x < 0.4$ and positive for $x > 0.4$ (there are other regions of interest).

Then, let us start Newton at $x=0.35$; the successive iterates are $0.209896, 0.223146, 0.223130$. Then, no overshoot.

Now, let us start Newton at $x=0.45$; the successive iterates are $0.126103, 0.231506, 0.223126, 0.223130$. Then, one overshoot at the first iteration.

Amzoti
  • 56,093
Claude Leibovici
  • 260,315