# in Fixed point iterations, how is it that the image of an image will get me to the fixed point?

I've been following a lesson on Fixed point iterations, but I'm having a trouble to intuitively understand how is it that continuously plugging in the image of an image will lead me to the fixed point given that the conditions of fixed point theorem are satisfied, which are :

1-f(x) is continuous on closed interval [a,b]

2- for all c ,where c belongs to [a,b], f(c) lies between [a,b]

3- g prime exists on (a, b)

4- a constant 0 < k < 1 exists with g prime(x) ≤ k, for all x ∈ (a, b).

the way I picture it is like a seesaw, where the value of x is at one end and y at the other. at a fixed point, the seesaw is in balanced state where x is equal to y, and based on the function say for example ,$${a*x^2 +c}$$, at left of the fixed point x < y, which if plugged in will get me to the right of the fixed point where x > y which will take me back to the left and so on, but why is it that wobble will slow down and eventually reach the balanced state where x=y or close it by a tolerance margin? why not just keep rocking back and forth?

• Please state the "fixed point theorem" that you are studying. It's hard to anwer your question without knowing the hypotheses and conclusions of the theorem you are thinking about. – Lee Mosher Aug 1 '16 at 16:53
• it's boils down to four conditions thatwill prove that a unique fixed point exists for a given interval. 1-f(x) is continuous on closed interval [a,b] 2- for all c ,where c belongs to [a,b], f(c) lies between [a,b] 3- g prime exists on (a, b) 4- a constant 0 < k < 1 exists with g prime(x) ≤ k, for all x ∈ (a, b). – Raed Tabani Aug 1 '16 at 16:58
• A couple of other things: you could improve your question still further by using MathJax for math formulas, and by putting the statement of the theorem into the question. – Lee Mosher Aug 1 '16 at 17:01
• The way I understand your "wobble" is an iteration of $f$ of the form $$x_1, \quad x_2 = f(x_1), \quad x_3 = f(x_2), \quad f_4 = f(x_3), ...$$ and a fixed point $p$ of $f$, such that $x_i$ is to the left of $p$ if $i$ is odd, whereas $x_i$ is to the right of $p$ of $i$ is even. Is this correct? – Lee Mosher Aug 1 '16 at 17:07
• exactly, it would alternate for one side of p if i is odd and to the other when even – Raed Tabani Aug 1 '16 at 17:15

You mention a hypothesis that there exists $0<k<1$ such that $g'(x) \le k$ for all $x \in (a,b)$. I think what you meant is that $|g'(x)| \le k$ for all $x \in (a,b)$ (without that, it is not possible for an orbit to switch back on forth on opposite sides of a fixed point).
You are asking about an iteration of the form $$x_1, \quad x_2 = f(x_1), \quad x_3 = f(x_2), \quad f_4 = f(x_3), ...$$ and a fixed point $p$ of $f$, such that $x_i$ is to the left of $p$ if $i$ is odd, whereas $x_i$ is to the right of $p$ of $i$ is even. So $p$ is between $x_i$ and $x_{i+1}$ (for each $i=1,2,3,...$). I will give a special proof in this situation that $f(x_i)$ must converge to $p$. It might be helpful for you to compare this special proof with the general proof (in your lesson?) that every iteration converges to $p$.
The key inequality needed is that $$|x_{i+1}-x_i| \le k \cdot |x_i-x_{i-1}|$$ In fact $|f(x)-f(y)| \le k \cdot |x-y|$ for all $x,y \in (a,b)$, which is an easy consequence of the mean value theorem and the hypothesis that $|g'(x)| \le k$.
It follows that the sequence of real numbers $$(*)\qquad |x_2-x_1|, \quad |x_3-x_2|, \quad |x_4-x_3|, \quad |x_5 - x_4|, \quad ...$$ is term-by-term less than the sequence $$(**)\qquad |x_2 - x_1|, \quad k \cdot |x_2 - x_1|, \quad k^2 \cdot |x_2 - x_1|, \quad k^3 \cdot |x_2 - x_1|, \quad ...$$ Since $0<k<1$, it follows that the sequence of numbers $(*)$ converges to zero. Since $p$ is between $x_i$ and $x_{i+1}$, by the triangle inequality it follows that $d(x_i,p) \le d(x_i,x_{i+1})$, and so $d(x_i,p)$ converges to zero, forcing the sequence $x_i$ to converge to $p$.
• It's not a question of whether the derivative is $<1$ or $>1$. Instead, it is a question of whether the derivative is positive or negative, that's why I added absolute value signs to the hypothesis. Suppose $x_1 < x_2$. If the derivative is positive then the function is increasing, and so $x_2 = f(x_1) < f(x_2) = x_3$. Continuing this argument, $x_1 < x_2 < x_3 < x_4 < ...$. – Lee Mosher Aug 1 '16 at 20:05