Newton's Method special case troubles....

• February 7th 2011, 04:00 PM
Glockstock
Newton's Method special case troubles....
Most analyses of Newton's method assume that $f'(a) \neq 0$,
where a is the solution of the nonlinear equation.
I'd like to investigate what happens in the special case where $f'(a) = 0$ but $f''(a) \neq 0$
I'd like to show that:
$\lim_{x \to a}{f(x) \over [f'(x)]^2} = {1 \over 2f''(a)}$
I'm pretty sure I need to use L'H^ospital's rule, but I haven't figured out how to make it work out...I keep getting the numerator to be zero, but I need to to be one.
After this is done, I have to show that in this case, the iteration function $g(x) = {x - {f(x) \over f'(x)}}$
generated by Newton's method is a contraction. Thanks in advance for any tips to get me going on this!
• February 7th 2011, 04:58 PM
Lord Voldemort
How can you use L'Hopital's rule in the limit?
You can't assume that f(a) = 0, which, because f'(a) = 0, must be true to use it. Otherwise your needed conclusion is trivial. Sorry that I can't help you... but maybe you're on the wrong track here?
• February 7th 2011, 05:10 PM
Glockstock
Quote:

Originally Posted by Lord Voldemort
How can you use L'Hopital's rule in the limit?
You can't assume that f(a) = 0, which, because f'(a) = 0, must be true to use it.

Well, our professor offered LHopital's rule as a hint for the problem. If you apply it to the limit, it yields:
$\lim {f(x)\over[f'(x)]^2} = \lim {f'(x)\over 2f''(x)}$
which looks similar to what I'm looking for, except for the pesky f'(x) in the numerator of the right hand side...
I may have left out the fact that we're assuming a is a solution of f(x)=0.
• February 7th 2011, 05:12 PM
Lord Voldemort
In that case, remember chain rule!
• February 7th 2011, 05:14 PM
Glockstock
Quote:

Originally Posted by Lord Voldemort
In that case, remember chain rule!

AHHHH totally overlooked that. Thank you very much he-who-shall-not-be-named.
• February 7th 2011, 05:16 PM
Lord Voldemort
Also I'm not sure about the second part... what is a contraction?
Perhaps I haven't done enough calculus yet to help you with this, but there ARE very advanced people that come on this forum if you wait long enough.
• February 7th 2011, 05:17 PM
Glockstock
But, now with the proof out of the way, I'm still having a bit of trouble proving the iteration function of Newton's method being a contraction...I hope I just am overlooking something trivial again.
• February 7th 2011, 05:18 PM
skeeter
Quote:

Originally Posted by Glockstock
Well, our professor offered LHopital's rule as a hint for the problem. If you apply it to the limit, it yields:
$\lim {f(x)\over[f'(x)]^2} = \lim {f'(x)\over 2f''(x)}$
which looks similar to what I'm looking for, except for the pesky f'(x) in the numerator of the right hand side...
I may have left out the fact that we're assuming a is a solution of f(x)=0.

that assumption makes it rather simple ...

given $f(a) = 0$ and $f'(a) = 0$

$\displaystyle \lim_{x \to a} \frac{f(x)}{[f'(x)]^2}$

L'Hopital (note use of the chain rule in the denominator)...

$\displaystyle \lim_{x \to a} \frac{f'(x)}{2 f'(x) \cdot f''(x)} =$

$\displaystyle \lim_{x \to a} \frac{1}{2f''(x)}$
• February 7th 2011, 05:20 PM
Glockstock
Quote:

Originally Posted by Lord Voldemort
Also I'm not sure about the second part... what is a contraction?
Perhaps I haven't done enough calculus yet to help you with this, but there ARE very advanced people that come on this forum if you wait long enough.

Perhaps, thanks for just looking into my trivial oversight. A contraction is defined as a function in the interval [a,b] where: f(x) - f(y) <= L(x - y). 0 <= L < 1. An alternate definition is: f'(x) <= L. I may have to just wait. Thanks again.
• February 7th 2011, 05:23 PM
Lord Voldemort
Deleted, one moment.
L is a constant between 0 inclusive and 1? Or a function?
• February 7th 2011, 05:28 PM
Glockstock
Quote:

Originally Posted by Lord Voldemort
Deleted, one moment.
L is a constant between 0 inclusive and 1? Or a function?

A constant.
• February 7th 2011, 05:36 PM
Glockstock
Ah, I've got it myself. A contraction is defined as $g'(x) < L$ where $0 <= L < 1$
so if $g(x) = {x - {f(x) \over f'(x)}}$ then $g'(x) = {f(x)f''(x) \over [f'(x)]^2}$.
When this is applied with the above equation, we know ${f(x) \over [f'(x)]^2} = {1 \over 2f''(x)}$. So applying this, with cancellations, $g'(x) = {1 \over 2}$. This is in the interval. Thanks!