# Fixed point problem

• May 23rd 2008, 08:46 PM
arbolis
Fixed point problem
The problem states : Let $f(x)=x^3+4x^2-10$. The equation $f(x)=0$ has a unique root $r$ approximately equals to $1.365230013$ in $[1,2]$.
Show that the following function has a fixed point in $r$ : $g(x)=\frac{\sqrt{10-x^3}}{2}$.
I'm trying to understand the fixed point signification since about 3 days but I really don't get the concept, and much less how to do exercises related to it. So if you could do this problem explaining, I will be very very grateful.
• May 23rd 2008, 08:52 PM
Isomorphism
Quote:

Originally Posted by arbolis
The problem states : Let $f(x)=x^3+4x^2-10$. The equation $f(x)=0$ has a unique root $r$ approximately equals to $1.365230013$ in $[1,2]$.
Show that the following function has a fixed point in $r$ : $g(x)=\frac{\sqrt{10-x^3}}{2}$.
I'm trying to understand the fixed point signification since about 3 days but I really don't get the concept, and much less how to do exercises related to it. So if you could do this problem explaining, I will be very very grateful.

This needs only the definition of fixed point... why are you getting stuck?

Essentially you are asking us to show there exists an x such that $g(x) = x$
$\frac{\sqrt{10-x^3}}{2} = x \Rightarrow 10 - x^3 = 4x^2 \Rightarrow x^3+4x^2-10 = 0$

But you have already found out a value for this! Yes, x=1.365230013 is the fixed point of g(x).
• May 23rd 2008, 09:06 PM
arbolis
Quote:

Essentially you are asking us to show there exists an x such that g(x)=x
I didn't know that. So all what I had to do is to replace $x$ by $r$ in $g(x)-x$?

Quote:

This needs only the definition of fixed point... why are you getting stuck?
Because I don't understand it well.
Quote:

But you have already found out a value for this! Yes, x=1.365230013 is the fixed point of g(x).
I didn't! It was a given information.
Thanks a lot, if I could I'd push twice on the thanks button!
Correct me if I'm wrong : a fixed point of a function is the point where the function cross the y=x line?
The b part of the exercise ask me to do 4 iterations using the fixed point method starting with $x_0=1.5$. I think I know how to do so, but I don't understand why it will converges to the root. Is this method fast, compared to the Newton's one?
And a big question : if $g(x)$ wasn't given, how could I have guessed a function that would work for the fixed point method? (I guess the method to find such a function is more or less as complicated as to approximate the root of g!)
• May 23rd 2008, 09:27 PM
Isomorphism
Quote:

Originally Posted by arbolis
I didn't know that. So all what I had to do is to replace $x$ by $r$ in $g(x)-x$?

What is r?(Thinking)

Quote:

Correct me if I'm wrong : a fixed point of a function is the point where the function cross the y=x line?
Yes :)

Quote:

The b part of the exercise ask me to do 4 iterations using the fixed point method starting with $x_0=1.5$. I think I know how to do so, but I don't understand why it will converges to the root. Is this method fast, compared to the Newton's one?
I dont know actually, since I have not studied these numerical methods. But Wiki says that Newtons method is equivalent to fixed point. I am not sure though but I am sure CaptainBlack can clarify. He is very good at such Numerical Algorithms.

Quote:

And a big question : if $g(x)$ wasn't given, how could I have guessed a function that would work for the fixed point method?
Just a thought: Cant you just take g(x) = f(x) - x ?
• May 23rd 2008, 09:39 PM
arbolis
To
Quote:

What is r?(Thinking)
Hmm, I take this as a yes, without understanding well! But I know that many functions can do the job, it's the case of g in the exercise, which is not $f(x)-x$.